Pls help urgently extra points and mark brainlist

What value for x will make the equation true?

Rama09 [41]

Answer:

C . 4

Step-by-step explanation:

step 1 : -3 multiply with inside of the ( 7x-5 )

= -21x + 15

step 2 : then it will be 21x + 15 = 87 + 3x

step 3 : switch between the odds into like this

21x - 3x = 87 - 15

step 4 : subtract it all

18x = 72

step 5 : bring 18 to the other side

x = 72/18

step 6 : to find the ans of x is just calculate them

x = 72 ÷ 18

= 4

Or you can do like these ⬇️

-3 ( 7x - 15 ) = 87 + 3x

21x + 15 = 87 + 3x

21x - 3x = 87 - 15

18x = 72

x = 72/18 or 72÷18

x = 4

4 0

3 years ago

The sugar content of the syrup in canned peaches is normally distributed. A random sample of n = 10 cans yields a sample standar

Serggg [28]

Answer:

The 95% confidence interval is given by:

3.30<σ<8.76

Step-by-step explanation:

1) Data given and notation

s=4.798 represent the sample standard deviation

$\bar x$ represent the sample mean

n=10 the sample size

Confidence=95% or 0.95

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=10-1=9$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a tabel to find the critical values.

The excel commands would be: "=CHISQ.INV(0.025,9)" "=CHISQ.INV(0.975,9)". so for this case the critical values are:

$\chi^2_{\alpha/2}=19.022$

$\chi^2_{1- \alpha/2}=2.700$

And replacing into the formula for the interval we got:

$\frac{(9)(4.798)^2}{19.022} \leq \sigma \frac{(9)(4.798)^2}{2.700}$

$10.892 \leq \sigma^2 \leq 76.736$

Now we just take square root on both sides of the interval and we got:

$3.30 \leq \sigma \leq 8.76$

So the best option would be:

3.30<σ<8.76

7 0

3 years ago

1. If a scale factor is applied to a figure and all dimensions are changed proportionally, what is the effect on the perimeter o

tester [92]

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

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

Perimeter

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

Area

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

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

Perimeter

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

Area

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

5 0

3 years ago

Graph the triangle with the following coordinates. Find the area of the triangle 2 different ways.

stira [4]

what triangle you have a picture?

8 0

3 years ago

a container of candy is shaped like a cylinder and has a volume of 125.6 cubic centimeters. If the heights of the container is 1

Elenna [48]

The radius of the container is 2 centimeter

<h3>Solution:</h3>

Given that a container of candy is shaped like a cylinder