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kakasveta [241]
2 years ago
5

Marcie wants to enclose her yard with a fence. Her yard is in the shape of a rectangle attached to a triangle. The formula for t

he area of the enclosed space is A = lw + 0.5bh. Solve for b .
a. b = A divided by the quantity l times w plus 0.5 times h B . b = The quantity A minus l times w all divided by 0.5 times h
c. b = A – lw – 0.5h
d. b = A + lw + 0.5h No Idea :o
Mathematics
2 answers:
nydimaria [60]2 years ago
8 0

Let

l------> the length side of the rectangle

w----> the width side of a rectangle

b----> the base of the triangle

h----> the height of the triangle

we know that

the area of the yard is equal to

A=lw+0.5bh

Solve for b------> ( that means clear variable b)

Subtract lw both sides

A-lw=lw+0.5bh-lw

A-lw=0.5bh

Divide by 0.5h both sides

(A-lw)/(0.5h)=0.5bh/(0.5h)

b=(A-lw)/(0.5h)

therefore

<u>the answer is the option B</u>

The quantity A minus l times w all divided by 0.5 times h


DerKrebs [107]2 years ago
8 0
A=lw+0.5bh \ \ \ \ \ \ \ \ \ \ |-lw \\&#10;A-lw=0.5bh \ \ \ \ \ \ \ \ \ \ |\div 0.5h \\&#10;\boxed{b=\frac{A-lw}{0.5h}}

The answer is B.
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Given:

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To find:

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Solution:

We have,

|x-4(72)|=2

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It can be written as

x-288=\pm 2

Add 288 on both sides.

x=288\pm 2

x=288-2 and x=288+2

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Therefore, the highest and lowest scores Sam could have made in the tournament are 290 and 286 respectively.

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A study is being conducted in which the health of two independent groups of ten policyholders is being monitored over a one-year
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Answer:

46.91% probability that at least nine participants complete the study in one of the two groups, but not in both groups

Step-by-step explanation:

We use two binomial trials 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.

Probability of at least nine participants finishing the study in a group.

0.2 probability of a students dropping out. So 1 - 0.2 = 0.8 probability of a student finishing the study. This means that p = 0.8.

10 students, so n = 10

We have to find:

P(X \geq 9) = P(X = 9) + P(X = 10)

Then

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

P(X = 9) = C_{10,9}.(0.8)^{9}.(0.2)^{1} = 0.2684

P(X = 10) = C_{10,10}.(0.8)^{10}.(0.2)^{0} = 0.1074

P(X \geq 9) = P(X = 9) + P(X = 10) = 0.2684 + 0.1074 = 0.3758

0.3758 probability that at least nine participants complete the study in a group.

Calculate the probability that at least nine participants complete the study in one of the two groups, but not in both groups?

0.3758 probability that at least nine participants complete the study in a group. This means that p = 0.3758

Two groups, so n = 2

We have to find P(X = 1).

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

P(X = 2) = C_{2,1}.(0.3758)^{1}.(0.6242)^{1} = 0.4691

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Answer:

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