What is the domain of the function y = StartRoot x EndRoot?

What is the Slope of the line passing through the points (0,4) and (-8, -1

Contact [7]

Slope = (y2 - y1) / (x2 - x1)
(0,4)...x1 = 0 and y1 = 4
(-8,-1)...x2 = -8 and y2 = -1
now we sub and solve
slope = (-1 - 4) / (-8 - 0) = -5/-8 = 5/8 <=

3 0

3 years ago

Read 2 more answers

−3=___+(−6) what is the missing number?

alexandr402 [8]

Answer:

-3

Step-by-step explanation:

-3 + -3 = -6

Hope This Helped

3 0

2 years ago

Read 2 more answers

Mark donated 15% of the money he earned last summer to the hurricane relief fund. If he donated at total of $475.00 how much did

Fantom [35]

That’s the right answer well done

4 0

3 years ago

Determine the values of x and y in the diagram and use them to determine the measures of the eight angles

AleksAgata [21]

Answer:

Part 1) $x=8$

Part 2) $y=12$

Part 3) $m\angle a=105^o$

Part 4) $m\angle b=75^o$

Part 5) $m\angle c=105^o$

Part 6) $m\angle d=75^o$

Part 7) $m\angle e=105^o$

Part 8) $m\angle f=75^o$

Part 9) $m\angle g=105^o$

Part 10) $m\angle h=75^o$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

Part 1) Determine the value of x

we know that

$m\angle a=m\angle g$ ----> by alternate exterior angles

substitute the given values

$(17x-31)^o=(12x+9)^o$

solve for x

$17x-12x=9+31\\5x=40\\x=8$

Part 2) Determine the value of y

we know that

$m\angle c+m\angle d=180^o$ ----> by supplementary angles (consecutive interior angles)

substitute the given values

$(7y+21)^o+(5y+15)^o=180^o$

solve for y

$(12y+36)^o=180^o$

$12y=180-36$

$12y=144$

$y=12$

Part 3) Find the measure of angle a

we have

$m\angle a=(17x-31)^o$

substitute the value of x

$m\angle a=(17(8)-31)=105^o$

Part 4) Find the measure of angle b

we know that

$m\angle a+m\angle b=180^o$ ----> by supplementary angles (consecutive interior angles)

we have

$m\angle a=105^o$

substitute

$105^o+m\angle b=180^o$

$m\angle b=180^o-105^o$

$m\angle b=75^o$

Part 5) Find the measure of angle c

we know that

$m\angle c=m\angle a$ ----> by vertical angles

we have

$m\angle a=105^o$

therefore

$m\angle c=105^o$

Part 6) Find the measure of angle d

we know that

$m\angle d=m\angle b$ ----> by vertical angles

we have

$m\angle b=75^o$

therefore

$m\angle d=75^o$

Part 7) Find the measure of angle e

we know that

$m\angle e=m\angle a$ ----> by corresponding angles

we have

$m\angle a=105^o$

therefore

$m\angle e=105^o$

Part 8) Find the measure of angle f

we know that

$m\angle f=m\angle b$ ----> by corresponding angles

we have

$m\angle b=75^o$

therefore

$m\angle f=75^o$

Part 9) Find the measure of angle g

we know that

$m\angle g=m\angle c$ ----> by corresponding angles

we have

$m\angle c=105^o$

therefore

$m\angle g=105^o$

Part 10) Find the measure of angle h

we know that

$m\angle h=m\angle d$ ----> by corresponding angles

we have

$m\angle d=75^o$

therefore

$m\angle h=75^o$

6 0

3 years ago

An educational psychologist wishes to know the mean number of words a third grader can read per minute. She wants to make an est

zhannawk [14.2K]

Answer:

99% confidence interval for the mean number of words a third grader can read per minute is [24.7 , 25.1].

Step-by-step explanation:

We are given that an educational psychologist wishes to know the mean number of words a third grader can read per minute. She wants to make an estimate at the 99% level of confidence. For a sample of 4657 third graders, the mean words per minute read was 24.9. Assume a population standard deviation of 5.7.

So, the pivotal quantity for 99% confidence interval for the mean number of words a third grader can read per minute is given by;

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

where, $\bar X$ = sample mean words per minute read = 24.9

$\sigma$ = population standard deviation = 5.7

n = sample of third graders = 4657

$\mu$ = population mean

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99

P(-2.5758 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

<u>99% confidence interval for </u> $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $24.9-2.5758 \times {\frac{5.7}{\sqrt{4657} } }$ , $24.9+2.5758 \times {\frac{5.7}{\sqrt{4657} } }$ ]

= [24.7 , 25.1]

Therefore, 99% confidence interval for the mean number of words a third grader can read per minute is [24.7 , 25.1].

6 0

3 years ago