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3 years ago

A dairy cow can produce 5400 quarts of milk per year. Suppose there are about 6.4 million cows in the U.S.

Mathematics

1 answer:

Ilya [14]3 years ago

5 0

Answer:

3.46 x $10^{10}$

Step-by-step explanation:

Number of cows 6.4 million = 6.4 x $10^{6}$

production rate for each cow per year = 5400 = 5.4 x 10³

Total amount of milk produced per year,

= 6.4 x $10^{6}$ x 5.4 x 10³

= (6.4) (5.4) x $10^{6+3}$

= 34.56 x $10^{9}$

= 3.46 x $10^{10}$

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3 years ago

A combined total of $46,000 is invested in two bonds that pay 4% and

valkas [14]

Answer: 4%-16000 and 9.5%-30000

Why:

1) x+y=46000

x=46000-y

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3 years ago

A public health official is planning for the supplyof influenza vaccine needed for the upcoming flu season. She wants to estimat

Marizza181 [45]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25$

And rounded up we have that n=421

Step-by-step explanation:

We know that the sample proportion have the following distribution:

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

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 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

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 $ME =\pm 0.04$ 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 assume that a prior estimation for p would be $\hat p =0.5$ since we don't have any other info provided. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25$

And rounded up we have that n=421

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3 years ago