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sveta [45]
3 years ago
11

A hotel manager is adding a tile border around the​ hotel's rectangular pool. Let p represent the width of the​ pool, in feet. T

he length is 3 more than 3 times the​ width, as shown. Write two expressions that represent the perimeter of the pool.

Mathematics
2 answers:
Ronch [10]3 years ago
6 0

Answer:

A and F

Step-by-step explanation:

Stells [14]3 years ago
3 0
It’s A&F, because when finding perimeter, you’re adding all of the sides together. Two of the sides are equal to “p”, while two of them are equal to “3p+3”, with this it would be “p+p+3p+3+3p+3”, and once simplified would result in “8p+6”.
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Around 80-75 percent.
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3 years ago
The table shows the relationship between the number that fit into given number of boxes from a bakery . What is the missing valu
ki77a [65]

Answer:

The missing value should be 48 because between the two numbers of each group. They are multplied by 6

Step-by-step explanation:

Hope this Helped!

7 0
2 years ago
Read 2 more answers
7 tenths times 10 is another problem that i need help on
Elza [17]
Hi there!

7 tenths times 10 is pretty easy and can be done on a calculator.
7 tenths is 0.7 in standard form.

0.7 × 10 = 7

Because if you think about it...
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8 0
3 years ago
A random sample of 100 automobile owners in thestate of Virginia shows that an automobile is driven onaverage 23,500 kilometers
Anastaziya [24]

Answer:

a) 22497.7 < μ< 24502.3

b)  With 99% confidence the possible error will not exceed 1002.3

Step-by-step explanation:

Given that:

Mean (μ) = 23500 kilometers per year

Standard deviation (σ) = 3900 kilometers

Confidence level (c) = 99% = 0.99

number of samples (n) = 100

a) α = 1 - c = 1 - 0.99 = 0.01

\frac{\alpha }{2} =\frac{0.01}{2}=0.005\\ z_{\frac{\alpha }{2}}=z_{0.005}=2.57

Using normal distribution table, z_{0.005 is the z value of 1 - 0.005 = 0.995 of the area to the right which is 2.57.

The margin of error (e) is given as:

e= z_{0.005}\frac{\sigma}{\sqrt{n} }  = 2.57*\frac{3900}{\sqrt{100} } =1002.3

The 99% confidence interval = (μ - e, μ + e) = (23500 - 1002.3, 23500 + 1002.3) =  (22497.7, 24502.3)

Confidence interval = 22497.7 < μ< 24502.3

b) With 99% confidence the possible error will not exceed 1002.3

5 0
3 years ago
The graph represents the normal distribution of recorded weights, in pounds, of cats at a veterinary clinic.
ICE Princess25 [194]

Answer:

The weights within this range are:

8.9

9.5

9.8

10.4


Step-by-step explanation:

The graph represents the normal distribution with a mean of 9.5 and a standard deviation of 0.5. Therefore, the range of the values within 2 standard deviations is:

9.5 - 2(0.5) ≤ x ≤ 9.5 + 2(0.5)

8.5 ≤ x ≤ 10.5



8 0
3 years ago
Read 2 more answers
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