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Nikolay [14]
3 years ago
7

It was recently estimated that homes without children outnumber homes with children by about 8 to 7. If there are 3870 homes in

a county, how many of them are homes without children?
Mathematics
1 answer:
damaskus [11]3 years ago
4 0

Answer: x = 2,340 homes without children

There are 2,340 homes without children.

Step-by-step explanation:

'x' represent the number of homes without children

'y' represent the number of homes with children

x : y = 9 : 7

Since the total number of homes is 4160 means x + y = 4160.

From x + y = 4160 follows y = 4160 - x.

Plug y = 4160 - x into x : y = 9 : 7

x : (4160 - x) = 9 : 7

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What are two inequalities for “at least 10” please answer .
sladkih [1.3K]

Answer:

the term "at least" means the least x is allowed to be is 10. therefore, x>10 and x≥10

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3 years ago
Milo wants to make a mixture that is 50% lemon juice and 50% lime juice. How much 100% lemon juice should he add to a juice mixt
Vanyuwa [196]

Answer:

1.5gal

Step-by-step explanation:

From the given information:

50% lemon juice + 50% lime juice is the intended mixture.

Therefore, to find quantity that would be added to 20% lemon juice and 80% lime juice, let's assume this to be x.

x of 100% and 4-x gallons of the existing 20% mix

100%x + 20%(4-x) = 50%(4)

1.00x + 0.2(4-x) = 0.5(4)

1.00x + 0.8-0.2x = 2

1.00x-0.2x = 2-0.8

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× = 1.5gal.

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4 years ago
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29/4 as a percentage
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5 0
4 years ago
Claim: Most adults would erase all of their personal information online if they could.A software firm survey of 453 randomly sel
Anit [1.1K]

Answer:

The value of the test statistic is 4.26.

Step-by-step explanation:

In this case a hypothesis test is performed to determine whether most adults would erase all of their personal information online if they could.

A random sample of <em>n</em> = 453 adults were selected and it was found that 60% of them would erase all of their personal information online if they could.

Assume that the population proportion is, <em>p</em> = 0.50.

A <em>z</em>-test for single proportion would to used to perform the test.

Compute the value of the test statistic as follows:

z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}

  =\frac{0.60-0.50}{\sqrt{\frac{0.50(1-0.50)}{453}}}\\\\=\frac{0.10}{0.0235}\\\\=4.25532\\\\\approx 4.26

Thus, the value of the test statistic is 4.26.

3 0
3 years ago
When the population distribution is normal, the statistic median {|X1 − X tilde|, . . . , |Xn − X tilde|}/0.6745 can be used to
Andreyy89

Answer:

The corresponding point estimate is 0.882.

The sample standard deviation is 1.373.

Step-by-step explanation:

The data set is:

S = {25.01, 25.87, 26.34, 26.51, 26.75, 27.24, 27.40, 27.63, 27.83, 27.90, 28.08, 28.13, 28.37, 28.58, 28.59, 28.96, 29.20, 29.22, 29.38, 30.88}

Compute the mean as follows:

\bar X=\frac{1}{n}\sum X\\\\=\frac{1}{20}\times [25.01+25.87+...+30.88]\\\\=\frac{1}{20}\times 557.87\\\\=27.8935

Subtract the mean from each value and take the modulus of those values.

The new data set is:

S₁ = {2.8835, 2.0235, 1.5535, 1.3835, 1.1435, 0.6535, 0.4935, 0.2635, 0.0635, 0.0065, 0.1865, 0.2365, 0.4765, 0.6865, 0.6965, 1.0665, 1.3065, 1.3265, 1.4865, 2.9865}

Arrange these values in ascending order as follows:

S₂ = {0.0065 , 0.0635 , 0.1865 , 0.2365 , 0.2635 , 0.4765 , 0.4935 , 0.6535 , 0.6865 , 0.6965 , 1.0665 , 1.1435 , 1.3065 , 1.3265 , 1.3835 , 1.4865 , 1.5535 , 2.0235 , 2.8835 , 2.9865}

There are 20 observations in the data set.

The median value for an even set of values is the mean of the middle two values.

In this case the median will be the mean of the 10th and 11th observations.

\text{Median}=\frac{10^{th}obs.+11^{th}obs.}{2}=\frac{0.6965+1.0665}{2}=0.8815\approx 0.882

Thus, the corresponding point estimate is 0.882.

Compute the standard deviation as follows:

In set S₁ we computed the absolute mean deviations.

Now take the square of these values and divide by (n - 1) to compute the sample variance:

\sigma^{2}=\frac{1}{n-1}\sum (|X_{i}-\bar X|)^{2}

     =\frac{1}{20-1}\times [(2.8835)^{2}+(2.0235)^{2}+...+(2.9865)^{2}]\\\\=\frac{1}{19}\times 35.7953\\\\=1.88396

Compute the sample standard deviation as follows:

\sigma=\sqrt{\sigma^{2}}=\sqrt{1.88396}=1.373

Thus, the sample standard deviation is 1.373.

6 0
3 years ago
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