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Virty [35]
3 years ago
6

How many nickels are in 25.2 dollars

Mathematics
2 answers:
frosja888 [35]3 years ago
7 0

1 nickel = 0.05

25.2/0.05 = 504 nickels

777dan777 [17]3 years ago
7 0
There should be 504 nickels in 25.2 dollars. Hope this was helpful.
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Answer:

Ah yes, one of my favorite topics.

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Step 2: First one equals to 1.125. Next, find the LCD (lowest common dominator, which is 10). \frac{9}{10} -\frac{6}{10}=\frac{3}{10}

Step 3: 1.125+3/10

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A survey of 90 men found that an average amount spent on St. Patrick's day of $55 with a standard deviation of $18. A similar su
dem82 [27]

Answer:

The value of the test statistic is 4.70.

Step-by-step explanation:

The hypothesis for this test can be defined as follows:

<em>H</em>₀: Men do not spend more than women on St. Patrick's day, i.e. μ₁ = μ₂.

<em>H</em>ₐ: Men spend more than women on St. Patrick's day, i.e. μ₁ > μ₂.

The population standard deviations are not known.

So a <em>t</em>-distribution will be used to perform the test.

The test statistic for the test of difference between mean is:

t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}}} }

Given:

\bar x_{1}=55\\s_{1}=18\\n_{1}=90\\\bar x_{1}=44\\s_{1}=16\\n_{1}=86

Compute the value of the test statistic as follows:

t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}}} }=\frac{55-44}{\sqrt{\frac{15^{2}}{90}+\frac{16^{2}}{86} }}=4.70

Thus, the value of the test statistic is 4.70.

7 0
3 years ago
at a summer camp 23 of the boys and 17 of the girls can swim there nine boys 11 girls at the camp who cannot swim. what is the r
Sedbober [7]

Answer:

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Step-by-step explanation:

looking at the last row, we can see that the total number of students will be 4×30+5×30=120+150=270

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2 years ago
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