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3 years ago

Question 10

Mathematics

1 answer:

mylen [45]3 years ago

8 0

-4，2

Step-by-step explanation:

find the average of the x coordinates and y coordinates

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Tina wants to save money for school. Tina invests $1000 in an account that pays an interest rate of 8%. How many years will it t

Neko [114]

1300 = 1000 (1.08)^t
1.3 = (1.08)^t
log base 1.08 (1.3) = t
3.41 = t

8 0

3 years ago

The heights of women in the USA are normally distributed with a mean of 64 inches and a standard deviation of 3 inches.

Rainbow [258]

Answer:

(a) 0.2061

(b) 0.2514

Step-by-step explanation:

Let <em>X</em> denote the heights of women in the USA.

It is provided that <em>X</em> follows a normal distribution with a mean of 64 inches and a standard deviation of 3 inches.

(a)

Compute the probability that the sample mean is greater than 63 inches as follows:

$P(\bar X>63)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{63-64}{3/\sqrt{6}})\\\\=P(Z>-0.82)\\\\=P(Z$

Thus, the probability that the sample mean is greater than 63 inches is 0.2061.

(b)

Compute the probability that a randomly selected woman is taller than 66 inches as follows:

$P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-64}{3})\\\\=P(Z>0.67)\\\\=1-P(Z$

Thus, the probability that a randomly selected woman is taller than 66 inches is 0.2514.

(c)

Compute the probability that the mean height of a random sample of 100 women is greater than 66 inches as follows:

$P(\bar X>66)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{66-64}{3/\sqrt{100}})\\\\=P(Z>6.67)\\\\\ =0$

Thus, the probability that the mean height of a random sample of 100 women is greater than 66 inches is 0.

8 0

3 years ago

Solve for x please and thank you

Sphinxa [80]

9514 1404 393

Answer:

x = 7

Step-by-step explanation:

The product of segment lengths is the same for the two chords.

9(21) = x(27)

x = 189/27 . . . . . divide by 27

x = 7

5 0

3 years ago

Miguel volunteers at his local food pantry and takes note of how much money is donated each day during a 10-day fundraising phon

tangare [24]

Answer:bbb

Step-by-step explanation:

4 0

3 years ago

I have <img src="https://tex.z-dn.net/?f=f%28t%29%3D%20%20%5Cfrac%7B%20t%5E%7B2%7D%20%7D%7B9%7D%20" id="TexFormula1" title="f(t)

skad [1K]

$\bf \begin{cases}
f(t)=\cfrac{t^2}{9}\\\\
g(t)=e^{\cfrac{}{}\frac{-ln(t)}{4}}\\\\
k(t)=f(t)+c\cdot g(t)
\end{cases}\implies k(t)=\cfrac{t^2}{9}+C\cdot e^{\cfrac{}{}\frac{-ln(t)}{4}}
\\\\\\
k(1)=4\implies 
\begin{cases}
k=4\\
t=1
\end{cases}\implies 4=\cfrac{1^2}{9}+C\cdot e^{\cfrac{}{}\frac{-ln(1)}{4}}
\\\\\\
4=\cfrac{1}{9}+C\cdot e^{\frac{0}{4}}\implies 4=\cfrac{1}{9}+C\cdot 1\implies 4-\cfrac{1}{9}=C
\\\\\\
\cfrac{35}{9}=C$

6 0

3 years ago