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2 years ago

8

Find the surface area for the triangular prism below. Whoever answers right the quickest will get brainliest. PLEASE HELP QUICKL

Y!!!

1 answer:

LenKa [72]2 years ago

6 0

Answer:

To answer your question use the code ICE on here to get your answer works every time for me hope this helps

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14+{-2+3[1+3(-6-2)]}

Marrrta [24]

14 + {-2 + 3[1 + 3(-6-2)]}
14 + { -2 + 3[1 + 3(-8)]}
14 + { -2 + 3[1 - 24]}
14 + { -2 + 3[-23]}
14 + {-2 - 69}
14 + {-71}
14 - 71
- 57 <==

3 0

3 years ago

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Fraction less than 1/2. 3/4 1/6 6/12 4/10 5/8 5/2

N76 [4]

Fractions less than 1/2 are 1/6 & 4/10.
Fractions greater than 1 are 5/4 & 15/12.
Fractions closer to 0 than to 1 are 1/8, 1/4, & 3/10.

5 0

3 years ago

5) Mario owns a plumbing business. He charges an hourly rate plus a onetime set up fee. His first job was 4 hours and he charged

Wittaler [7]

Answer:

Results are below.

Step-by-step explanation:

<u>First, we need to calculate the variable income per job:</u>

<u />

First job= 450 - 150= 300

Second job= 675 - 150= 525

<u>Now, the hourly rate:</u>

<u></u>

Hourly rate= total variable income / number of hours

First job= 300 / 4= $75

Second job= 525 / 3= $175

It is probably that the second job was 7 hours long. If not, he doesn't charge the same amount per hour

5 0

3 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

I am Lyosha [343]

Answer:

(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

Step-by-step explanation:

Let <em>X</em> = number of students who read above the eighth grade level.

(a)

A sample of <em>n</em> = 269 students are selected. Of these 269 students, <em>X</em> = 224 students who can read above the eighth grade level.

Compute the proportion of students who can read above the eighth grade level as follows:

$\hat p=\frac{X}{n}=\frac{224}{269}=0.8327$

The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

$1-\hat p=1-0.8327$

$=0.1673$

Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

<em>n</em> = 709

<em>X</em> = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

$\hat q=1-\hat p$

$=1-\frac{X}{n}$

$=1-\frac{546}{709}$

$=0.2299\\\approx 0.229$

The critical value of <em>z</em> for 95% confidence interval is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the 95% confidence interval for the population proportion as follows:

$CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$=0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)$

Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

5 0

3 years ago

What is 0.325 written as a percent?

bezimeni [28]

Answer:

32.5%

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers