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2 years ago

Domain and Range Question

Mathematics

2 answers:

svetoff [14.1K]2 years ago

5 0

The answer is a. hope that helped

mote1985 [20]2 years ago

5 0

The third answer is the correct one.

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A mailbox that is 36 inches tall is beside a tree.The length of the mailboxes shadow is 28 inches.The length of the trees shadow

mixer [17]

Answer:

10.5 feet

Step-by-step explanation:

In this problem, we have two similar triangles:

- One consists of the mailbox, its shadow and the hypothenuse

- The other one consists of the tree, its shadow and the hypothenuse

The two triangles are similar, so they have same proportions between their sides: therefore, we can apply the rule of three:

$\frac{m}{s_m}=\frac{t}{s_t}$

where

m = 36 in is the height of the mailbox

$s_m=28 in$ is the shadow of the mailbox

t is the height of the tree

$s_t=98 in$ is the length of the shadow of the tree

Solving for t, we find the height of the tree:

$t=\frac{m\cdot s_t}{s_m}=\frac{(36)(98)}{28}=126 in$

And since

1 feet = 12 inches

The height of the tree in feet is

$t=\frac{126}{12}=10.5 ft$

3 0

2 years ago

Simplify this expression: a + 2a + 3b - a - b

aniked [119]

Simplified would be 2a+2b

8 0

2 years ago

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A marketing research company desires to know the mean consumption of meat per week among people over age 30. A sample of 2092 pe

prisoha [69]

Answer:

95% confidence interval for the mean consumption of meat among people over age 30 is [2.9 pounds , 3.1 pounds].

Step-by-step explanation:

We are given that a sample of 2092 people over age 30 was drawn and the mean meat consumption was 3 pounds. Assume that the population standard deviation is known to be 1.4 pounds.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean meat consumption = 3 pounds

s = population standard deviation = 1.4 pounds

n = sample of people = 2092

$\mu$ = population mean consumption of meat

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $3-1.96 \times {\frac{1.4}{\sqrt{2092} } }$ , $3+1.96 \times {\frac{1.4}{\sqrt{2092} } }$ ]

= [2.9 pounds , 3.1 pounds]

Therefore, 95% confidence interval for the mean consumption of meat among people over age 30 is [2.9 pounds , 3.1 pounds].

7 0

2 years ago

Write the equation of the line that passes through the points (4,7) and (1,9) in point-slope form.

Nesterboy [21]

$(\stackrel{x_1}{4}~,~\stackrel{y_1}{7})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{9}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{9}-\stackrel{y1}{7}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{4}}} \implies \cfrac{9 -7}{1 -4} \implies \cfrac{ 2 }{ -3 } \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{7}=\stackrel{m}{-\cfrac{2}{3}}(x-\stackrel{x_1}{4})$

6 0

1 year ago

A bag contains 2 red marbles and 3 black marbles. If Abby picks a marble without looking, returns it to the bag and then draws a

Morgarella [4.7K]

Answer:

Therefore, the probability that both marbles are red would be 4 over25.

Step-by-step explanation:

4 0

2 years ago

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