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Vikentia [17]
3 years ago
10

What is the equation of the line that passes through the point (-3, 4) and has a slope of 1

Mathematics
1 answer:
VMariaS [17]3 years ago
3 0
The equation is: y = x + 7
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The value of x is 88° enjoy

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What is the area of a triangle that has a
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A. 45 in

Step-by-step explanation:

To find area you multiply length and width so you would multiply 9 and 10 but since it is a triangle you would divide by two

Hope this helps!

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Luke earned a salary of 48,048 last year. how much did he earn per month?
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1. If a scale factor is applied to a figure and all dimensions are changed proportionally, what is the effect on the perimeter o
scZoUnD [109]

Answer:

Part 1) The perimeter of the new figure must be equal to the perimeter of the original figure multiplied by the scale factor (see the explanation)

Part 2) The area of the new figure must be equal to the area of the original figure multiplied by the scale factor squared

Part 3) The new figure and the original figure are not similar figures (see the explanation)

Step-by-step explanation:

Part 1) If a scale factor is applied to a figure and all dimensions are changed proportionally, what is the effect on the perimeter of the figure?

we know that

If all dimensions are changed proportionally, then the new figure and the original figure are similar

When two figures are similar, the ratio of its perimeters is equal to the scale factor

so

The perimeter of the new figure must be equal to the perimeter of the original figure multiplied by the scale factor

Part 2) If a scale factor is applied to a figure and all dimensions are changed proportionally, what is the effect on the area of the figure?

we know that

If all dimensions are changed proportionally, then the new figure and the original figure are similar

When two figures are similar, the ratio of its areas is equal to the scale factor squared

so

The area of the new figure must be equal to the area of the original figure multiplied by the scale factor squared

Part 3) What would happen to the perimeter and area of a figure if the dimensions were changed NON-proportionally? For example, if the length of a rectangle was tripled, but the  width did not change? Or if the length was tripled and the width was decreased by a factor of 1/4?​

we know that

If the dimensions were changed NON-proportionally, then the ratio of the corresponding sides of the new figure and the original figure are not proportional

That means

The new figure and the original figure are not similar figures

therefore

Corresponding sides are not proportional and corresponding angles are not congruent

so

<em>A) If the length of a rectangle was tripled, but the  width did not change?</em>

<em>Perimeter</em>

The original perimeter is P=2L+2W

The new perimeter would be P=2(3L)+2W ----> P=6L+2W

The perimeter of the new figure is greater than the perimeter of the original figure but are not proportionals

<u>Area</u>

The original area is A=LW

The new area  would be A=(3L)(W) ----> A=3LW

The area of the new figure is three times the area of the original figure but its ratio is not equal to the scale factor squared, because there is no single scale factor

<em>B) If the length was tripled and the width was decreased by a factor of 1/4?</em>

<u>Perimeter</u>

The original perimeter is P=2L+2W

The new perimeter would be P=2(3L)+2(W/4) ----> P=6L+W/2

The perimeter of the new figure and the perimeter of the original figure are not proportionals

<u><em>Area</em></u>

The original area is A=LW

The new area  would be A=(3L)(W/4) ----> A=(3/4)LW

The area of the new figure is three-fourth times the area of the original figure but its ratio is not equal to the scale factor squared, because there is no single scale factor

4 0
3 years ago
Historically, a certain region has experienced 92 thunder days annually. (A "thunder day" is day on which at least one instance
DiKsa [7]

Answer:

We conclude that the mean number of thunder days is less than 92.

Step-by-step explanation:

We are given that Historically, a certain region has experienced 92 thunder days annually.

Over the past fifteen years, the mean number of thunder days is 72 with a standard deviation of 38.

<u><em>Let </em></u>\mu<u><em> = mean number of thunder days.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 92 days     {means that the mean number of thunder days is more than or equal to 92}

Alternate Hypothesis, H_A : \mu < 92 days     {means that the mean number of thunder days is less than 92}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

                       T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean number of thunder days = 72

             s = sample standard deviation = 38

            n = sample of years = 15

So, <u><em>test statistics</em></u>  =  \frac{72-92}{\frac{38}{\sqrt{15} } }  ~ t_1_4

                               =  -2.038

The value of z test statistics is -2.038.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical value of -1.761 at 14 degree of freedom for left-tailed test.

Since our test statistics is less than the critical value of t as -2.038 < -1.761, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the mean number of thunder days is less than 92.

7 0
3 years ago
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