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Snezhnost [94]
3 years ago
14

On the distant planet Cowabunga, the weights of cows have a normal distribution with a mean of 479 pounds and a standard deviati

on of 40 pounds. The cow transport truck holds 15 cows and can hold a maximum weight of 7350. If 15 cows are randomly selected from the very large herd to go on the truck, what is the probability their total weight will be over the maximum allowed of 7350
Mathematics
1 answer:
Eddi Din [679]3 years ago
5 0

Answer: 0.1435

Step-by-step explanation:

Given : Mean = 479 pounds

Standard deviation = 40 pounds.

Let  X denote the weights of cows.

X\sim N(\mu=479,\sigma=40)

The cow transport truck holds 15 cows and can hold a maximum weight of 7350.

i.e. mean weight of cow in this case =\overline{x}=\dfrac{7350}{15}=490\text{ pounds}

If 15 cows are randomly selected from the very large herd to go on the truck, what is the probability their total weight will be over the maximum allowed of 7350 will be:-

P(\overline{x}>490)=P(\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{490-479}{\dfrac{40}{\sqrt{15}}})\\\\=P(z>1.065)\ \ [\because\ z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=1-P(z\leq1.065)\\\\=1- 0.8565=0.1435\ \ [\text{By z-table}]

Hence, If 15 cows are randomly selected from the very large herd to go on the truck,  the probability their total weight will be over the maximum allowed of 7350 = 0.1435

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How many weeks of data must be randomly sampled to estimate the mean weekly sales of a new line of athletic footwear? We want 98
Irina18 [472]

Answer:

n=(\frac{2.33(1400)}{400})^2 =66.50 \approx 67

So the answer for this case would be n=67 rounded up

Step-by-step explanation:

Information given

\bar X represent the sample mean for the sample  

\mu population mean

\sigma=1400 represent the sample standard deviation

n represent the sample size  

Solution to the problem

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =400 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

The critical value for 98% of confidence interval now can be founded using the normal distribution. And the critical value would be z_{\alpha/2}=2.33, replacing into formula (b) we got:

n=(\frac{2.33(1400)}{400})^2 =66.50 \approx 67

So the answer for this case would be n=67 rounded up

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3 years ago
Regis leans a 10-foot ladder against a wall. The base of the ladder makes a 65 angle with the ground.
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The distance, In feet, from the base of the ladder to the base of the wall is 4.2 ft.

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The situation forms a right angle triangle.

<h3>Right angle triangle</h3>

Right angle triangle has one of its angles as 90 degrees. The sides and angle can be found using trigonometric ratios.

The length of the ladder is the hypotenuse of the triangle formed. Therefore, the distance, In feet, from the base of the ladder to the base of the wall can be calculated as follows;

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Therefore, he needs to move the ladder 0.1 ft closer to the building.

learn more on right angle triangle here: brainly.com/question/14988069

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Step-by-step explanation:

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