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Wittaler [7]
3 years ago
8

a person made 4 blankets each blanket used 180 inches of ribbon for the border how many yards of ribbon did the person use for a

ll 4 blankets
Mathematics
2 answers:
adell [148]3 years ago
4 0

Answer:


Step-by-step explanation:


Elden [556K]3 years ago
4 0

Answer:

20 yards of ribbon were used to make 4 blankets.

Step-by-step explanation:

36 inches = 1 yard.

180 inches * 4 = 720 inches.

720/36 = 20 yards.

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How many solutions are there to the system of equations? 4x-5y=5 and -.08x+.10y= 0.10No solution 1 solution 2 solutionsInfinite
dsp73

Infinite solutions

hope this helps

4 0
4 years ago
Your bedroom wall is 221 inches wide. You want to hang two posters, each 28 inches across, with an equal space between each post
olga2289 [7]

Answer:

55 inches

Step-by-step explanation:

2 posters x 28 inches each = 56

221 - 56 = 165 inches

3 blank spaces on wall

165 ÷ 3 = 55 inches between posters and on either side

5 0
3 years ago
A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. ​(a) What must be true
valentina_108 [34]

Answer:

(a) The distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b) The value of P(\bar X is 0.7642.

(c) The value of P(\bar X\geq 69.1) is 0.3670.

Step-by-step explanation:

A random sample of size <em>n</em> = 10 is selected from a population.

Let the population be made up of the random variable <em>X</em>.

The mean and standard deviation of <em>X</em> are:

\mu=68\\\sigma=15

(a)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. <em>n</em> = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

\mu_{\bar x}=\mu=68

And the standard deviation of the distribution of the sample mean is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74

Thus, the distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b)

Compute the value of P(\bar X as follows:

P(\bar X

                    =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the value of P(\bar X is 0.7642.

(c)

Compute the value of P(\bar X\geq 69.1) as follows:

Apply continuity correction as follows:

P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)

                    =P(\bar X>69.6)

                    =P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})

                    =P(Z>0.34)\\=1-P(Z

Thus, the value of P(\bar X\geq 69.1) is 0.3670.

7 0
3 years ago
Find the least common denominator (lcd) of 1/3 and 2/9 .
OLEGan [10]
The lowest common denominator of 1/3 and 2/9 = 9
8 0
3 years ago
Find the sum of 3.8 + (-2.9)
MakcuM [25]

Answer:

3.8 - 2.9= .9

ANSWER : .9

6 0
3 years ago
Read 2 more answers
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