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3 years ago

8

At noon, Sarah headed from Elk City to Isabel at 60 miles per hour. Two hours later, Joan grade from Isabel to Elk City at 46 mi

les per hour. If it is 332 miles from Elk City to Isabel, what time did they meet. Solve this using this format

1 answer:

navik [9.2K]3 years ago

8 0

I thank that 46 + 60= 106 miles per hour
106 - 332= 226 miles per hour
Explain: 106+ 226 = 332 miles per hour

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She can put the rocks into,
2 groups of 19
Or one group of 38.

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Age If you add Natalie's age and Fred's age, the result is 38. If you add Fred's age to 4 times

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Solve the linear equation: <br><br> <img src="https://tex.z-dn.net/?f=4%5E%7B2x%2B7%7D%20%3D%208%5E%7B2x-3%7D" id="TexFormula1"

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Answer:

x = 11.5

Step-by-step explanation:

Taking the logarithm base 2 will transform this to a linear equation.

2(2x+7) = 3(2x -3)

0 = 3(2x -3) -2(2x +7) . . . . subtract the left side

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8 0

3 years ago

A data set with a mean of 34 and a standard deviation of 2.5 is normally distributed

tresset_1 [31]

Answer:

a) $z= \frac{34-34}{2.5}= 0$

$z= \frac{39-34}{2.5}= 2$

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

b) $P(X$

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So one deviation below the mean we have: (100-68)/2 = 16%

c) $z= \frac{29-34}{2.5}= -2$

$z= \frac{36.5-34}{2.5}= 1$

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

Step-by-step explanation:

For this case we have a random variable with the following parameters:

$X \sim N(\mu = 34, \sigma=2.5)$

From the empirical rule we know that within one deviation from the mean we have 68% of the values, within two deviations we have 95% and within 3 deviations we have 99.7% of the data.

We want to find the following probability:

$P(34 < X$

We can find the number of deviation from the mean with the z score formula:

$z= \frac{X -\mu}{\sigma}$

And replacing we got

$z= \frac{34-34}{2.5}= 0$

$z= \frac{39-34}{2.5}= 2$

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

For the second case:

$P(X$

$z= \frac{31.5-34}{2.5}= -1$

So one deviation below the mean we have: (100-68)/2 = 16%

For the third case:

$P(29 < X$

And replacing we got:

$z= \frac{29-34}{2.5}= -2$

$z= \frac{36.5-34}{2.5}= 1$

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

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3 years ago

One endpoint of a line segment is (–6, 4), and the midpoint of the segment is (–2, 7). What are the coordinates of the other end

Arturiano [62]

Answer:

A. (2, 10)

Step-by-step explanation:

If there are two points (x1,y1) and (x2,y2) on a coordinates plane then

coordinates of the midpoint is given by (x1+x2)/2, (y1+y2)/2

__________________________________________________

Given

one endpoint is (–6, 4) and midpoint is (–2, 7)

let coordinates of other endpoint be (x,y)

now if one endpoint is (–6, 4) and other endpoint is (x,y)

then coordinates of midpoint is ((-6+x)/2 , (4+y)/2)

while the coordinates of midpoint given is (–2, 7)

thus (–2, 7) should be equal to ((-6+x)/2 , (4+y)/2)

-2 = (-6+x)/2 and 7 = (4+y)/2

=> -2*2 = -6+x => 7*2 = 4+y

=> -4 = -6+x => 14 = 4+y

=> -4+6 = x => 14-4 = y

=> x = 2 => 10 = y

Thus, coordinates of the other endpoint is(2,10)

6 0

3 years ago