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3 years ago

11

if our monthly income is $1,740 and your house payment is $1,253 what fraction of your monthly income must go to pay your house

payment

1 answer:

Lady_Fox [76]3 years ago

7 0

5/6 ? maybe and it might be 1253/1740

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6. The area of a circle is 36pi cm squared. Find its circumference.

Mumz [18]

Answer:

AREA=PI*6*6 THUS THE RADIUS IS 6 CM. C=37.68 CM.

Step-by-step explanation:

3 0

3 years ago

PLZZZZ HELP The Lakeview city council is considering two different plans to help reduce the amount of pollution in the city. Pla

sdas [7]

1. The two equations follow the slope-intercept form which y=mx + b. They have the same y-intercept which is equal to 900.

2. Suppose x= 10
Plan Alpha: y=-7x+900=-7(10)+900=830
Plan Foxtrot: y = -17x+900 = -17(10)+900=730

Between the 2 plans, the lower value is 730. Plan Foxtrot is better for reduction of pollution.

3. The city council should choose Plan Foxtrot because the reduction of pollution is much greater. Their goal would be reached faster if they go with faster reduction of pollution.

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3 years ago

Given d(t) = -2t^2 - 3t + 13, what is the

Aloiza [94]

Answer:

I'm pretty sure b is the answer

4 0

3 years ago

Why did the cow keep jumping over the barrel

Artemon [7]

Hi , the answer for this riddle is "she was practicing for a moonshot".

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3 years ago

A normal curve with a mean of 500 and a standard deviation of 100 is shown. Shade the region under the curve within one standard

N76 [4]

Answer:

And for this case we want to find how much of the data within the shaded region given in the figure attached.

And we can use the zscore formula given by:

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

And if we find the z score for the limits we got:

$z = \frac{400-500}{100}=-1$

$z = \frac{600-500}{100}= 1$

And we want to find this probability using the normal standard distribution or excel:

$P(-1$

So we would expect about 68.2 % of the data within one deviation from the mean

Step-by-step explanation:

For this case we know that the variable of interest let's say x follows a normal distribution with the following parameters:

$X \sim N(\mu = 500, \sigma =100)$

And for this case we want to find how much of the data within the shaded region given in the figure attached.

And we can use the zscore formula given by:

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

And if we find the z score for the limits we got:

$z = \frac{400-500}{100}=-1$

$z = \frac{600-500}{100}= 1$

And we want to find this probability using the normal standard distribution or excel:

$P(-1$

So we would expect about 68.2 % of the data within one deviation from the mean

6 0

4 years ago