All categories

4 years ago

Pleaseeeeeeeeeeeee help :)

Mathematics

1 answer:

mash [69]4 years ago

6 0

The table says that the x value represents the number the years since 1985. To find the number of bald eagle pairs in 1998, we need to calculate x:

1998 - 1985 = 13

Now plug 13 in for x to find y:

y = 312.5(13) + 1562.5

y = 5625

You might be interested in

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.7

Zinaida [17]

Answer:

(a) The probability that the sample average sediment density is at most 3.00 is 0.092.

The probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b) The sample size must be at least 77.

Step-by-step explanation:

The random variable X ca be defined as the sediment density (g/cm) of a specimen from a certain region.

The random variable X is normally distributed with mean, μ = 2.7 and standard deviation, σ = 0.75.

(a)

A random sample of n = 25 specimens is selected.

Compute the probability that the sample average sediment density is at most 3.00 as follows:

Apply continuity correction:

$P(\bar X\leq 3.00)=P(\bar X$

$=P(\bar X$

$=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}$

$=P(Z$

Thus, the probability that the sample average sediment density is at most 3.00 is 0.092.

Compute the probability that the sample average sediment density is between 2.70 and 3.00 as follows:

$P(2.70$

$=P(0$

*Use a z-table.

Thus, the probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b)

It is provided that:

$P(\bar X\leq 3.00)\geq 0.99$

$P(\bar X$

$P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}$

$P(Z$

The value of z for the probability above is

z ≥ 2.33

Compute the value of n as follows:

$\frac{|2.50-2.70|}{0.75/\sqrt{n}}\geq 2.33$

$\frac{|-0.20|}{2.33}\geq \frac{0.75}{\sqrt{n}}$

$\sqrt{n}\geq \frac{0.75\times 2.33}{|-0.20|}\\\sqrt{n}\geq 8.7375\\n\geq 76.3439\\\approx n\geq 77$

Thus, the sample size must be at least 77.

8 0

3 years ago

(a^3b)^7(a^7b^7) simplify

Helen [10]

Answer:

Step-by-step explanation:

$\text{Use}\\\\(ab)^n=a^nb^n\\\\\bigg(a^n\bigg)^m=a^{nm}\\\\a^n\cdot a^m=a^{n+m}\\--------------------\\\\\bigg(a^3b\bigg)^7\bigg(a^7b^7\bigg)=\bigg(a^3\bigg)^7(b^7)(a^7)(b^7)=(a^{3\cdot7})(a^7)(b^7)(b^7)\\\\=\bigg(a^{21+7}\bigg)\bigg(b^{7+7}\bigg)=\boxed{a^{28}b^{14}}$