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2 years ago

Find the surface area of the net below.

Mathematics

1 answer:

dedylja [7]2 years ago

5 0

Answer:

472 in²

Step-by-step explanation:

This net is made up of 3 identical rectangles and 2 identical triangles.

Area of a rectangle = width x length

Area of a triangle = 1/2 x base x height

<u>Rectangle</u>

width = 8 in
length = 19 in

<u>Triangle</u>

base = 8 in
height = 2 in

<u>Surface Area</u>

SA = 3(8 x 19) + 2(0.5 x 8 x 2)

⇒ SA = 456 + 16

⇒ SA = 472 in²

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There is 4/6 of a pizza left from dinner . Each person gets 1/6 of a pizza for lunch the next day . How many people eat the pizz

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The number of people which gets 1/6 of the pizza as required for lunch the next day is; 4 persons.

<h3>What is the number of people who get 1/6 of the pizza for lunch?</h3>

It follows from the task content that only 4/6 of the pizza is left and each individual needs 1/6 of the pizza for lunch.

On this note, the 4/6 fraction of pizza left mean: 4 × 1/6.

Therefore, only four persons get lunch.

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2 years ago

Graph the inequality y>-4x+6 to represent its solution set.

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Step-by-step explanation:

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3 years ago

Solve for m.<br><br> -5m = -35<br> m = ?

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6 0

3 years ago

Read 2 more answers

Suppose that 50% of all babies born in a particular hospital are boys. If 6 babies born in the hospital are randomly selected, w

WINSTONCH [101]

Using the binomial distribution, it is found that there is a 0.3438 = 34.38% probability that fewer than 3 of them are boys.

<h3>What is the binomial distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem, the values of the parameters are given as follows:

n = 6, p = 0.5.

The probability that fewer than 3 of them are boys is given by:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

In which:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.5)^{0}.(0.5)^{6} = 0.0156$

$P(X = 1) = C_{6,1}.(0.5)^{1}.(0.5)^{5} = 0.0938$

$P(X = 2) = C_{6,2}.(0.5)^{2}.(0.5)^{4} = 0.2344$

Then:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0156 + 0.0938 + 0.2344 = 0.3438$

0.3438 = 34.38% probability that fewer than 3 of them are boys.

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7 0

2 years ago