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A cereal manufacturer is concerned that the boxes of cereal not be under filled or overfilled. Each box of cereal is supposed to

Vaselesa [24]

Answer:

We conclude that the population average weight of a cereal box is equal to 13 ounces.

Step-by-step explanation:

We are given that a cereal manufacturer is concerned that the boxes of cereal not be under filled or overfilled.

A random sample of 36 boxes is tested. The sample average weight is 12.85 ounces and the sample standard deviation is 0.75 ounces.

Let $\mu$ = population average weight of a cereal box

So, Null Hypothesis, $H_0$ : $\mu$ = 13 ounces {means that the population average weight of a cereal box is equal to 13 ounces}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 13 ounces {means that the population average weight of a cereal box differs from 13 ounces}

The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average weight = 12.85 ounces

$\sigma$ = sample standard deviation = 0.75 ounces

n = sample of boxes = 36

So, test statistics = $\frac{12.85-13}{\frac{0.75}{\sqrt{36} } }$ ~ $t_3_5$

= -1.20

The value of the test statistics is -1.20.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now at 5% significance level, the t table gives critical values between -2.03 and 2.03 at 35 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the population average weight of a cereal box is equal to 13 ounces.

3 0

2 years ago

A plane flying horizontally at an altitude of 2 mi and a speed of 540 mi/h passes directly over a radar station. Find the rate a

Rasek [7]

Answer:

The plane's distance from the radar station will increase about 8 miles per minute when it is 5 miles away from it.

Step-by-step explanation:

When the plane passes over the radar station, the current distance is the altitude h = 2. Then it moves b horizontally so that the distance to the station is 5. We can form a rectangle triangle using b, h and the hypotenuse 5. Therefore, b should satisfy

h²+b² = 5², since h = 2, h² = 4, as a result

b² = 25-4 = 21, thus

b = √21.

Since it moved √21 mi, then the time passed is √21/540 = 0.008466 hours, which is 0.51 minutes. Note that in 1 minute, the plane makes 540/60 = 9 miles.

The distance between the plane and the radar station after x minutes from the moment that the plane passes over it is given by the function

$f(x) = \sqrt{((9x)^2 + 2^2)} = \sqrt{(81x^2+4)}$

We have to compute the derivate of f in x = 0.51. The derivate of f is given by

$f'(x) = \frac{162x}{2\sqrt{81x^2+4}}$

also,

$f'(0.51) = \frac{162*0.51}{2\sqrt{81*0.51^2+4}} = 8.2486$

The plane's distance from the station will increase about 8 miles per minute.

4 0

3 years ago

(100 POINTS) Instructions:

erma4kov [3.2K]

Answer:

Length: 15

Width: 12

Height: 13

Step-by-step explanation:

Hope this helps!

(Mark as brainleist plz)

4 0

2 years ago

Problem

Paraphin [41]

Answer:

45/126 or 5/14

Step-by-step explanation:

a / b = 5/8

b / c = 9/14

a = 5

b = 8

b = 9

8 * 9 = 72

b = 72

If b was = 8 and now = 72, multiply a by 9.

b = 45

If b was 9 and now is 72, multiply c by 9.

c = 14 * 9 = 140 - 14 = 126

b / c = 45/126 or 5/14

You are welcome!

Kayden Kohl

9th Grade Student

4 0

1 year ago

I need your help because I don’t understand

kogti [31]

Alright, so first divide 20 km by 1.2 cm

16.6666666667

Multiply 16.6666666667 by 9

150

So it's 150 km!

Hope this helps!

5 0

3 years ago

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