The maximum acceptable level of a certain toxic chemical in vegetables has been set at 0.4
parts per million (ppm). A consumer health group measured the level of the chemical in a
random sample of tomatoes obtained from one producer.
The levels, in ppm, are shown below.
0.31 0.47 0.19 0.72 0.56
0.91 0.29 0.83 0.49 0.28
0.31 0.46 0.25 0.34 0.17
0.58 0.19 0.26 0.47 0.81
Does the data provide sufficient evidence to support the claim that the mean level of the
chemical in tomatoes from this producer is greater than the recommended level of 0.4 ppm?
Use a 0.05 significance level to test the claim that these sample levels come from a population
with a mean greater than 0.4 ppm. Use the P-value method of testing hypotheses. Assume that
the standard deviation of levels of the chemical in all such tomatoes is 0.21 ppm.
Choose the correct null and alternative hypotheses associated with this experiment.

Base on the equation, the amount Laura pay for 2 tennis balls and 4 golf balls $8

Sam buys 4 tennis balls and 12 golf balls for a total of £19.

let

x = price of each tennis ball

y = price of each golf ball

Hence,

4x + 12y = 19

Luke buys 6 tennis ball and 10 golf balls for a total of £22.50. Hence,

6x + 10y = 22.50

Combine the equation

4x + 12y = 19

6x + 10y = 22.50

6x + 18y = 28.5

6x + 10y = 22.50

8y = 6

y = 6 / 8

y = $0.75

6x + 10(0.75) = 22.50

6x + 7.5 = 22.50

6x = 22.50 - 7.5

6x = 15

x = 15 / 6

x = $2.5

The cost when Laura pays for 2 tennis balls and 4 golf balls is as follows:

2(2.5) + 4(0.75) = 5 + 3 = $8

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Answer:

r=14/2

r=7

[tex]circumference = 2\pi {r}\\ = 2 \times \frac{22}{7} \times 7 \\ = 44[/tex]

[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]

Let's find the circumference of given circle ~

Circumference :

[tex] \qquad \tt  \dashrightarrow \: \pi{d}^{} [/tex]

where, d = diameter

Now, let's calculate it ~

[tex] \qquad \tt  \dashrightarrow \: c = 14\pi{}^{} [/tex]

That's the answer I terms of π, but we can also plug in the approximate value of pi as 3.14

[tex] \qquad \tt  \dashrightarrow \: c = 14 \times 3.14{}^{} [/tex]

[tex] \qquad \tt  \dashrightarrow \: c = 43.96 \: \: m[/tex]

[tex] \qquad \tt  \dashrightarrow \: c \approx 44 \: \: m[/tex]

Answer:

113.04cm^2

Step-by-step explanation:

The area of a circle is pi * r^2

pi=3.14

So 6^2 is 36

36*3.14=113.04

Hope it helps!

Answer: -1

Step-by-step explanation:

Answer:

im thinking that is 3/74 because i count them

Step-by-step explanation:

3/74 rhe answer but 3/77 are wrong he just thinking

Answer:

3 over 77, 3/77

Step-by-step explanation:

You add up all of the marbles. (including the blue ones)

30+28+13+6 = 77.

So. You have a 3/77 chance of pulling a blue marble.

Answer:

x is equal to -34/16

Step-by-step explanation:

Hey there!

In order to solve for x, we need to first simplify everything

After simplifying, it will look like this:

[tex](4x^2+14x+16)-7x+14=4x^2-4-9x[/tex]

Now you can simplify by combining like terms

[tex]4x^2+7x+30=4x^2-9x-4[/tex]

You can cross out [tex]4x^2[/tex] from both sides and then simplify even more by again combining like terms

[tex]16x+34=0\\16x=-34\\x=-34/16[/tex]

So x is equal to -34/16

To solve the given equation, let's simplify and expand both sides:

[tex]\displaystyle\sf (2x+1)(2x+6)-7(x-2)=4(x+1)(x-1)-9x[/tex]

Expanding the brackets:

[tex]\displaystyle\sf (4x^{2}+13x+6)-7x+14=4(x^{2}-1)-9x[/tex]

Simplifying further:

[tex]\displaystyle\sf 4x^{2}+13x+6-7x+14=4x^{2}-4-9x[/tex]

Combining like terms:

[tex]\displaystyle\sf 4x^{2}+6x+20=4x^{2}-9x-4[/tex]

Subtracting [tex]\displaystyle\sf 4x^{2}[/tex] from both sides:

[tex]\displaystyle\sf 6x+20=-9x-4[/tex]

Bringing all the variables to one side and the constants to the other side:

[tex]\displaystyle\sf 6x+9x=-4-20[/tex]

[tex]\displaystyle\sf 15x=-24[/tex]

Dividing both sides by [tex]\displaystyle\sf 15[/tex] to solve for [tex]\displaystyle\sf x[/tex]:

[tex]\displaystyle\sf x=-\frac{24}{15}[/tex]

Simplifying the fraction:

[tex]\displaystyle\sf x=-\frac{8}{5}[/tex]

Therefore, the solution to the equation is [tex]\displaystyle\sf x=-\frac{8}{5}[/tex].

[tex]\huge{\mathfrak{\colorbox{black}{\textcolor{lime}{I\:hope\:this\:helps\:!\:\:}}}}[/tex]

♥️ [tex]\large{\underline{\textcolor{red}{\mathcal{SUMIT\:\:ROY\:\:(:\:\:}}}}[/tex]

Answer:

n=23

Step-by-step explanation:

a right angle measures 90

67+n=90

subtract 67 on both sides

n=23

Use angle addition postulate:

Substitute values to get the required equation and solve it for n:

Area of the yellow sector is 58.91 cm2

The spinner has a circular shape and would be solved as a circle.

Area of a circle = pi r^2

= 3.142 * 5^2

= 78.55 cm2

The circle is divided into 8

78.55 / 8 = 9.8188

Area of 6 parts = 6 * 9.8188
= 58.9128 cm2

Step-by-step explanation:

Given :-

To find:-

Solution :-

Formula for perimeter of rectangular swamp = 2(L+W)

124 ft. = 2( 5x -10 + x )

124 ft. = 2(6x - 10)

124 ft. = 12x - 20

124+20 = 12x

144/12 = x

12 = x

putting the values of x

Length = 5x-10

width = X

perimeter =2L+2W or 2(L+W)

124= 2(5W-10+W)

perimeter =2( 6W-10)

124 = 12W-20

124+20=(12W+20-20

144=12W

144÷12=12W÷12

12=W

Answer:

The answer is $1.85

Step-by-step explanation:

Now, we know that the cost of 5 chocolate bars are $5.75.

For 1 chocolate bar

5.75 ÷ 5 = $1.15

Here, we get the price of 1 chocolate bar

So, the cost of 2 chocolate bars and 3 packets of sweets are $7.85.

2 × chocolate bar = 2 × (1.15) = $2.3

Now, we want to find only the cost of one packet of sweets.

So,

7.85 – 2.3 = $5.55

3 packets of sweets cost $5.55

For 1 packets of sweets

5.55 ÷ 3 = $1.85

Thus, The cost of one packet of sweets = $1.85

Answer:

Step-by-step explanation:

Opposite angles of a cyclic quadrilateral sum to 180 so

m < x = 180-105 = 75 degrees

m < y = 180 - 95 = 85 degrees.

The apportion of Bronx quota is the alloted profit to Bronx

The apportion for Bronx quota is 20

The dataset is given as:

From the dataset, we have:

Bronx: 20.73

Remove the numbers after the decimal point

Bronx =  20

Hence, the apportion for Bronx quota is 20

Read more about quota at:

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Answer:

3/4

Step-by-step explanation:

Give brainliest if it helped...(:

5. Let [tex]x = \sin(\theta)[/tex]. Note that we want this variable change to be reversible, so we tacitly assume 0 ≤ θ ≤ π/2. Then

[tex]\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - x^2}[/tex]

and [tex]dx = \cos(\theta) \, d\theta[/tex]. So the integral transforms to

[tex]\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = \int \frac{\sin^3(\theta)}{\cos(\theta)} \cos(\theta) \, d\theta = \int \sin^3(\theta) \, d\theta[/tex]

Reduce the power by writing

[tex]\sin^3(\theta) = \sin(\theta) \sin^2(\theta) = \sin(\theta) (1 - \cos^2(\theta))[/tex]

Now let [tex]y = \cos(\theta)[/tex], so that [tex]dy = -\sin(\theta) \, d\theta[/tex]. Then

[tex]\displaystyle \int \sin(\theta) (1-\cos^2(\theta)) \, d\theta = - \int (1-y^2) \, dy = -y + \frac13 y^3 + C[/tex]

Replace the variable to get the antiderivative back in terms of x and we have

[tex]\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\cos(\theta) + \frac13 \cos^3(\theta) + C[/tex]

[tex]\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\sqrt{1-x^2} + \frac13 \left(\sqrt{1-x^2}\right)^3 + C[/tex]

[tex]\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\frac13 \sqrt{1-x^2} \left(3 - \left(\sqrt{1-x^2}\right)^2\right) + C[/tex]

[tex]\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = \boxed{-\frac13 \sqrt{1-x^2} (2+x^2) + C}[/tex]

6. Let [tex]x = 3\tan(\theta)[/tex] and [tex]dx=3\sec^2(\theta)\,d\theta[/tex]. It follows that

[tex]\cos(\theta) = \dfrac1{\sec(\theta)} = \dfrac1{\sqrt{1+\tan^2(\theta)}} = \dfrac3{\sqrt{9+x^2}}[/tex]

since, like in the previous integral, under this reversible variable change we assume -π/2 < θ < π/2. Over this interval, sec(θ) is positive.

Now,

[tex]\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = \int \frac{27\tan^3(\theta)}{\sqrt{9+9\tan^2(\theta)}} 3\sec^2(\theta) \, d\theta = 27 \int \frac{\tan^3(\theta) \sec^2(\theta)}{\sqrt{1+\tan^2(\theta)}} \, d\theta[/tex]

The denominator reduces to

[tex]\sqrt{1+\tan^2(\theta)} = \sqrt{\sec^2(\theta)} = |\sec(\theta)| = \sec(\theta)[/tex]

and so

[tex]\displaystyle 27 \int \tan^3(\theta) \sec(\theta) \, d\theta = 27 \int \frac{\sin^3(\theta)}{\cos^4(\theta)} \, d\theta[/tex]

Rewrite sin³(θ) just like before,

[tex]\displaystyle 27 \int \frac{\sin(\theta) (1-\cos^2(\theta))}{\cos^4(\theta)} \, d\theta[/tex]

and substitute [tex]y=\cos(\theta)[/tex] again to get

[tex]\displaystyle -27 \int \frac{1-y^2}{y^4} \, dy = 27 \int \left(\frac1{y^2} - \frac1{y^4}\right) \, dy = 27 \left(\frac1{3y^3} - \frac1y\right) + C[/tex]

Put everything back in terms of x :

[tex]\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = 9 \left(\frac1{\cos^3(\theta)} - \frac3{\cos(\theta)}\right) + C[/tex]

[tex]\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = 9 \left(\frac{\left(\sqrt{9+x^2}\right)^3}{27} - \sqrt{9+x^2}\right) + C[/tex]

[tex]\displaystyle \int \frac{x^3}{\sqrt{9+x^2}} \, dx = \boxed{\frac13 \sqrt{9+x^2} (x^2 - 18) + C}[/tex]

2(b). For some constants a, b, c, and d, we have

[tex]\dfrac1{x^2+x^4} = \dfrac1{x^2(1+x^2)} = \boxed{\dfrac ax + \dfrac b{x^2} + \dfrac{cx+d}{x^2+1}}[/tex]

3(a). For some constants a, b, and c,

[tex]\dfrac{x^2+4}{x^3-3x^2+2x} = \dfrac{x^2+4}{x(x-1)(x-2)} = \boxed{\dfrac ax + \dfrac b{x-1} + \dfrac c{x-2}}[/tex]

5(a). For some constants a-f,

[tex]\dfrac{x^5+1}{(x^2-x)(x^4+2x^2+1)} = \dfrac{x^5+1}{x(x-1)(x+1)(x^2+1)^2} \\\\ = \dfrac{x^4 - x^3 + x^2 - x + 1}{x(x-1)(x^2+1)^2} = \boxed{\dfrac ax + \dfrac b{x-1} + \dfrac{cx+d}{x^2+1} + \dfrac{ex+f}{(x^2+1)^2}}[/tex]

where we use the sum-of-5th-powers identity,

[tex]a^5 + b^5 = (a+b) (a^4-a^3b+a^2b^2-ab^3+b^4)[/tex]

Answer:

The distance is 20 units

Step-by-step explanation:

Answer:

7k + 6

Steps:

Subtract the numbers 20+6k-14k+k

6+6k+k

Combine the terms:

6+6k+k

6+7k

Rearange terms:

6+7k

7k+6

Answer:

-60p -6

Step-by-step explanation:

8(1 - 6p) - 2(6p + 7)

Distribute

8 - 48p  -12p -14

Combine like terms

-60p -6

Answer:

A. X < 5

Step-by-step explanation:

6(x -2) + 21 < 39

6x - 12 + 21 < 39

Group like terms

6x < 39 + 12 - 21

6x < 30

Divide through by 6

X < 30/6

X < 5

Answer:

When you graph the equations, both equations represent the same line. If a system has no solution, it is said to be inconsistent . The graphs of the lines do not intersect, so the graphs are parallel and there is no solution.

1320

Step-by-step explanation:

the triangle,1/2×15×20×2=300 .rectangle,17×25=425,20×17=340,15×17=255. sum them up 300+255+425+340=1320

bought price= 2000

sold price= 2500

the percentage profit

[tex]profit\% = 100 \times \frac{(final - initial)}{initial} [/tex]

[tex]profit\% = 100 \times \frac{(2500 - 2000)}{2000} [/tex]

[tex] = \frac{(2500 - 2000)}{2000} [/tex]

[tex] = \frac{1}{4} [/tex]

[tex] = \frac{1}{4} \times 100\%[/tex]

[tex]profit\% = 25\%[/tex]

therefore, his percentage profit is 25.

Answer: Its negetive. The answer is -52d^2

Hope this helped you!

Answer:

Step-by-step explanation:

[tex]-12\ d^{2}+-40\ d^{2} = -12\ d^{2}-\ 40\ d^{2} = -52\ d^{2}[/tex]

Since −52 is negative and d² is always positive for any number d

Then ,their product −52 d² is negative .

Answer:

I can help!! answer in image

Step-by-step explanation:

Have a great day sir

Step-by-step explanation:

A = 0.5(πr^2 )

A = 0.5(π × (20^2 ))

A = 628.32 units^2

Answer:

628.3 ft

Step-by-step explanation:

area of half circle = [tex]\frac{\pi r^{2}}{2}[/tex]

                             = [tex]\frac{\pi *20^{2} }{2}[/tex]

                             = 200[tex]\pi[/tex] = 628.3 ft

Answer:

it is very high because

Step-by-step explanation:

5-2+3 so maybe it a subtraction problem hope this works