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3 years ago

PLEASE HELP

Mathematics

1 answer:

pshichka [43]3 years ago

8 0

Answer:

1044 cm²

Step-by-step explanation:

Split the shape up into two

Area of shape A:

Area = length × width

To calculate the width = 46 - 21 = 25 cm

Area = 30 cm × 25 cm = 750cm²

Area of shape B

Area = length × width

Area = 14 cm × 21 cm = 294 cm²

Add up both of the areas:

750 cm² + 294 cm² = 1044 cm²

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Step-by-step explanation:

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3 years ago

A sociologist wants to know the opinions of employed adult women about government funding for day care. The women were asked to

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Answer:

The correct option is C 20.1%.

Step-by-step explanation:

The probability distribution chart representing the opinions of employed adult women about government funding for day care is as follows:

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4 years ago

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Question 4 of 10, Step 1 of 1 1/out of 10 Correct Certify Completion Icon Tries remaining:0 The Magazine Mass Marketing Company

erastovalidia [21]

Answer:

0.0105 = 1.05% probability that no more than 3 of the entry forms will include an order.

Step-by-step explanation:

For each entry form, there are only two possible outcomes. Either it includes an order, or it does not. The probability of an entry including an order is independent of any other entry, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

The Magazine Mass Marketing Company has received 16 entries in its latest sweepstakes.

This means that $n = 16$

They know that the probability of receiving a magazine subscription order with an entry form is 0.5.

This means that $p = 0.5$

What is the probability that no more than 3 of the entry forms will include an order?

At most 3 including an order, which is:

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

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

$P(X = 0) = C_{16,0}.(0.5)^{0}.(0.5)^{16} \approx 0$

$P(X = 1) = C_{16,1}.(0.5)^{1}.(0.5)^{15} = 0.0002$

$P(X = 2) = C_{16,2}.(0.5)^{2}.(0.5)^{14} = 0.0018$

$P(X = 3) = C_{16,3}.(0.5)^{3}.(0.5)^{13} = 0.0085$

Then

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3 years ago

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Shtirlitz [24]

Answer:

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Step-by-step explanation:

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So if we create a right triangle with that point after graphing it. We see the height of that triangle is 5 because that is the imaginary part. The base of that triangle has length $5\sqrt{3}$ . The problem is this doesn't give us any part of the angle we want, but it does give us the complementary of the part of the angle that is in second quadrant.