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3 years ago

Use the diagram to complete each statement. Twelve lines are drawn to form a cube as shown. Plane AFH is parallel to plane ?

Mathematics

1 answer:

Goshia [24]3 years ago

3 0

Answer:

BGC

Step-by-step explanation:

im pretty aure aorry if u get it wrong man

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George was weighed at the doctors office. The scale read 67.20 pounds. The doctor wrote 67.2 on George’s chart. Did the doctor m

nata0808 [166]

No because 67.20 rounded to the nearest tenth is 67.2

8 0

3 years ago

Y=-3x + 6 y=9 What is the solution to the system of equations

ASHA 777 [7]

Answer:

the answer is x=-1 and y=9. Hope this helps!

PLEASE GIVE BRAINLEST

7 0

3 years ago

Antonio is building a 1:240 scale model of a real castle. His model has a rectangular base that is 2.5 feet wide and 2 feet long

34kurt

The right approach to this problem is to first determine the real dimension by using the given scale factor.
width : (2.5 ft) x 250 = 625 ft
length: (2.0 ft) x 250 = 500 ft
The area of a rectangle is calculated by multiplying the length and the width.
A = (625 ft)(500 ft) ]= 312,500 ft²
Thus, the area of the actual castle in square feet is 312,500.

8 0

3 years ago

Read 2 more answers

It currently takes users a mean of 25 minutes to install the most popular computer program made by RodeTech, a software design c

ArbitrLikvidat [17]

Testing the hypothesis in this problem which is a two-tailed test, we can conclude that there is not sufficient evidence to conclude that the mean time of installation has changed, since the p-value of the test is 0.0364 > 0.01,

At the null hypothesis, we test if the mean is of 25 minutes, that is:

$H_0: \mu = 25$

At the alternative hypothesis, we test if the mean has changed, that is, if it is different than 25 minutes.

$H_1: \mu \neq 25$

We have the standard deviation for the sample, thus, the t-distribution is used. The value of the test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which:

X is the sample mean.
$\mu$ is the value tested at the null hypothesis.
s is the standard deviation of the sample.
n is the sample size.

In this problem:

25 is tested at the null hypothesis, thus $\mu = 25$ .
Sample mean of 26.2 minutes, thus $X = 26.2$ .
Sample of 51, thus $n = 51$ .
Variance of 16, thus $s = \sqrt{16} = 4$ .

The value of the test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{26.2 - 25}{\frac{4}{\sqrt{51}}}$

$t = 2.14$

The p-value of the test is found using a two-tailed test(test if the mean is different of a value), with t = 2.14 and 50 degrees of freedom.

Using a t-distribution calculator, this p-value is of 0.0364.

Since the p-value of the test is 0.0364 > 0.01, there is not sufficient evidence to conclude that the mean time of installation has changed.

A similar problem is given at brainly.com/question/23777908

3 0

3 years ago

Help, please. This is related to mirror and shadow problems

nadya68 [22]

Answer:

1. x/2=1.75/3.5 x=1

2. 4/2=y/3.5 y=3.5

3. P (Figure Left)=3+5+2+1+4+8+3+2=28 cm

A (Figure Left)=8x2-2x1-3x2=8 cm2

4. P (Figure Left) / P (Figure Right) = 1/1.75 = 28/ P (Figure Right)

P (Figure Right)= 49 cm

A (Figure Left) / A (Figure Right) = 1/1.752 = 8/ A (Figure Right)

A(Figure Right)= 24.5 cm2

Step-by-step explanation:

1. x/2=1.75/3.5 x=1

2. 4/2=y/3.5 y=3.5

3. P (Figure Left)=3+5+2+1+4+8+3+2=28 cm

A (Figure Left)=8x2-2x1-3x2=8 cm2

4. P (Figure Left) / P (Figure Right) = 1/1.75 = 28/ P (Figure Right)

P (Figure Right)= 49 cm

A (Figure Left) / A (Figure Right) = 1/1.752 = 8/ A (Figure Right)

A(Figure Right)= 24.5 cm2

5 0

3 years ago