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3 years ago

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A rectangle is shown. 2 right triangles sit on the top of the rectangle. The sides of the rectangle are congruent with 2 sides o

f the rectangle and have length x.
Which expression can be used to find the area of the composite figure?

4x
2x + 2x2
x2 + 2x
3x2

2 answers:

Salsk061 [2.6K]3 years ago

6 0

Answer:3x2

Step-by-step explanation:

Agata [3.3K]3 years ago

6 0

Answer:

$3x^{2}$

Step-by-step explanation:

just did this on edge

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A reporter for a student newspaper is writing an article on the cost of off-campus housing. A sample was selected of 10 one-bedr

olga55 [171]

Answer:

(507.05, 592.95)

Step-by-step explanation:

Given data:

sample mean = $550, sample standard deviation S = $60.05

95% confidence interval , n = 10

For 95% confidence interval for the mean

mean ± M.E.

where M.E. is margin of error = $t_{n-1}, \alpha/2\times\frac{S}{\sqrt{n} }$

Substituting the values in above equation

$=t_{10-1}, 0.05/2\times\frac{60.05}{\sqrt{10} }$

= 2.62×18.99

=42.955

= 550±42.95

=(507.05, 592.95)

6 0

3 years ago

One member of the debate team is going to be chosen president. each member is equally likely to be chosen. the probability that

inessss [21]

Let
x-------> <span>the probability that a boy is chosen
B------> </span>number of boys in the debate team
G------> number of girls in the debate team<span>

we know that
B+G=20-----> G=20-B-----> equation 1
</span>probability that a boy is chosen=number of boys in the debate team/total of people
so
x=B/20-------> equation 2

t<span>he probability that a girl is chosen is 2/3 the probability that a boy is chosen
</span>probability that a girl is chosen=(2/3)*x
probability that a girl is chosen=G/20
(2/3)*x=G/20-----> equation 3
substitute equation 1 in equation 3
(2/3)*x=[20-B]/20----> equation 4

resolve the equations system
x=B/20
(2/3)*x=[20-B]/20

using a graph tool
see the attached figure
the solution is
x=0.6
B=12
G=20-12-----> G=8

number of boys in the debate team is 12
number of girls in the debate team is 8
the probability that a boy is chosen is 0.6 ----> 60%
the probability that a girl is chosen is 0.6*(2/3)----> 0.4---> 40%

the answer is
number of girls in the debate team is 8

7 0

3 years ago

The graph shows the locations of a ball player seconds after

Katarina [22]

There is no graph. If you submit a graph I will answer this for you though.

4 0

3 years ago

Read 2 more answers

The length of the path of the first swing of the bob of a pendulum is 0.6 in. Each

ycow [4]

Answer:

0.3188646

Step-by-step explanation:

an=a×r^n-1

a6=0.6×0.9^5

=0.354294

1st swing: 0.6

2nd swing: 0.9×0.6=0.54

3rd swing: 0.9×0.54= 0.486

4th 0.9×0.486= 0.4374

5th 0.9×0.4374= 0.39366

6th 0.9×0.39366=0.354294

Brainliest please~

8 0

3 years ago

The systolic blood pressure (BP) of adults in the USA is nearly normally distributed with a mean of 121 and standard deviation o

Wittaler [7]

Answer:

Around 0.73% of adults in the USA have stage 2 high blood pressure

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 121 and standard deviation of 16.

This means that $\mu = 121, \sigma = 16$

Around what percentage of adults in the USA have stage 2 high blood pressure

The proportion is 1 subtracted by the p-value of Z when X = 160. So

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

$Z = \frac{160 - 121}{16}$

$Z = 2.44$

$Z = 2.44$ has a p-value of 0.9927.

1 - 0.9927 = 0.0073

0.0073*100% = 0.73%

Around 0.73% of adults in the USA have stage 2 high blood pressure

5 0

3 years ago