Look at picture************^^^^

What is the surface area of the right trapezoidal prism? To receive credit, you must show the work used to arrive at a final ans

julia-pushkina [17]

Answer:

210 cm²

Step-by-step explanation:

The net of the right trapezoidal prism consists of 2 trapezoid base and four rectangles.

Surface area of the trapezoidal prism = 2(area of trapezoid base) + area of the 4 rectangles

✔️Area of the 2 trapezoid bases:

Area = 2(½(a + b)×h)

Where,

a = 7 cm

b = 11 cm

h = 3 cm

Plug in the values

Area = 2(½(7 + 11)×3)

= (18 × 3)

Area of the 2 trapezoid bases = 54 cm²

✔️Area of Rectangle 1:

Length = 6 cm

Width = 3 cm

Area = 6 × 3 = 18 cm²

✔️Area of Rectangle 2:

Length = 7 cm

Width = 6 cm

Area = 7 × 6 = 42 cm²

✔️Area of Rectangle 3:

Length = 6 cm

Width = 5 cm

Area = 6 × 5 = 30 cm²

✔️Area of Rectangle 4:

Length = 11 cm

Width = 6 cm

Area = 11 × 6 = 66 cm²

✅Surface area of the trapezoidal prism = 54 + 18 + 42 + 30 + 66 = 210 cm²

7 0

2 years ago

Time spent using e-mail per session is normally distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that

liq [111]

Answer:

a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 11, \sigma = 3$

a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 25, s = \frac{3}{\sqrt{25}} = 0.6$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{11.2 - 11}{0.6}$

$Z = 0.33$

$Z = 0.33$ has a pvalue of 0.6293.

X = 10.8

$Z = \frac{X - \mu}{s}$

$Z = \frac{10.8 - 11}{0.6}$

$Z = -0.33$

$Z = -0.33$ has a pvalue of 0.3707.

0.6293 - 0.3707 = 0.2586

0.259 probability, rounded to three decimal places.

b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

X = 11

$Z = \frac{X - \mu}{s}$

$Z = \frac{11 - 11}{0.6}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

X = 10.5

$Z = \frac{X - \mu}{s}$

$Z = \frac{10.5 - 11}{0.6}$

$Z = -0.83$

$Z = -0.83$ has a pvalue of 0.2033.

0.5 - 0.2033 = 0.2967

0.297, rounded to three decimal places.

c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 100, s = \frac{3}{\sqrt{100}} = 0.3$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{11.2 - 11}{0.3}$

$Z = 0.67$

$Z = 0.67$ has a pvalue of 0.7486.

X = 10.8

$Z = \frac{X - \mu}{s}$

$Z = \frac{10.8 - 11}{0.3}$

$Z = -0.67$

$Z = -0.67$ has a pvalue of 0.2514.

0.7486 - 0.2514 = 0.4972

0.497, rounded to three decimal places.

5 0

3 years ago

A person borrows $50000 loan from bank at a rate of 10% for 5 years compounded yearly.

Hatshy [7]

<h3>Given:</h3>

P= $50,000
R= 10%
T= 5 years

P= Principal amount
R= Rate of interest
T= Time period

<h3>Solution:</h3>

$\large\boxed{Formula: A= P(1+ \frac{R}{100}{)}^{T}}$

Let's substitute according to the formula.

$A= 50000(1+ \frac{10}{100}{)}^{5}$

A= $80525.5

Now, we can find the interest paid

$\large\boxed{I= A-P}$

We'll have to deduct the total amount from the principal amount.

Let's substitute according to the formula.

$I= 80525.5-50000$

I= $30525.5

Hence, the total amount paid after 5 years is $80525.5 and $30525.5 was paid as interest.

7 0

2 years ago

describe and correct the error a student made finding the intercepts of the graph of the line 4x-6y=12

Kazeer [188]

Answer:

sorry????

Step-by-step explanation:

mmmmmmmmm?????

8 0

3 years ago

Ms. Wood took m1/6 of an hour to review each question fromt the Science test. How many questions can she review in 2/3 of an hou

Archy [21]

Answer: 4 questions

Step-by-step explanation:

16h=10 min

2/3=40 min

Use equation:

1:10=x:40

40=10x

x=40/10

x=4

5 0

2 years ago