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Kipish [7]
3 years ago
14

How much of each snack should hunter purchase to satisfy this senerio

Mathematics
1 answer:
bagirrra123 [75]3 years ago
3 0
B is the correct answer
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12x+7&lt;−11 OR 5x−8&gt;40<br> solve for x
kipiarov [429]

Answer:

x < -1.5 U x > 9.6

Step-by-step explanation:

12x + 7 < -11

x < -1.5

5x - 8 > 40

5x > 48

x > 9.6

6 0
3 years ago
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What is 32 divided by 147?
liq [111]
The answer is: 0.21768707483
7 0
3 years ago
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Sample Size for Proportion As a manufacturer of golf equipment, the Spalding Corporation wants to estimate the proportion of gol
Dima020 [189]

Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.025}{2.58})^2}=2662.56  

And rounded up we have that n=2663

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.025 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

We can assume an estimated proportion of \hat p =0.5 since we don't have prior info provided. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.025}{2.58})^2}=2662.56  

And rounded up we have that n=2663

6 0
3 years ago
Ashton answer this what is 8x8
yaroslaw [1]

Answer:

64

Step-by-step explanation:

3 0
3 years ago
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Adult panda weights are normally distributed with a mean of 200 pounds and a standard deviation of 20 pounds. The largest pandas
earnstyle [38]
\mathbb P(X>250)=1-\mathbb P(X\le250)

\mathbb P(X\le250)=\mathbb P\left(\dfrac{X-200}{20}\le\dfrac{250-200}{20}\right)=\mathbb P(Z\le2.5)\approx0.9938

which means

\mathbb P(X>250)\approx1-0.9938=0.0062

so approximately 0.62% of pandas weight over 250 pounds.
7 0
3 years ago
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