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How do I simplify a fraction? For example if the question is 7 1/5 divided by 2 2/15 what would I do to simplify?

Westkost [7]

7 1/5:
Multiply the denominator and the whole number: 7*5 = 35 and then add the numerator: 35+1= 36/5. Then, do this for the following fraction, the improper fraction value of 2 2/15 is 32/15.
Now we need 36/5 & 32/15 to have common denominators so that we can simplify it a little less complicated.

For Instance,

36/5*3= 108/15

Now we divide, (108/15)/(32/15)
108/32= 3.375 = 3 3/8

Answer is 3 3/8.

Hope this was easy to understand!

3 0

3 years ago

The residual plot for a data set is shown.

Cerrena [4.2K]

Answer:

The regression line is not a good model because there is a pattern in the residual plot.

Step-by-step explanation:

Given is a residual plot for a data set

The residual plot shows scatter plot of x and y

The plotting of points show that there is not likely to be a linear trend of relation between the two variables. It is more likely to be parabolic or exponential.

Hence the regression line cannot be a good model as they do not approach 0.

Also there is not a pattern of linear trend.

D) The regression line is not a good model because there is a pattern in the residual plot.

3 0

3 years ago

Read 2 more answers

40 tickets were sold for a concert, some at 75 cents and the rest at $1.25. if the total raised was $42, how many had the cheape

Murljashka [212]

Answer:

34

Step-by-step explanation:

$1.25 = 125 cents.

$42 = 4200 cents

Tickets sold at 75 cents = x

Tickets sold at 125 cents = y

x + y = 40

75x + 125y = 4200

Multiply the first equation by 75

75x + 75y = 3000

75x + 125y = 4200

Subtract the the second equation from the first.

75x + 75y = 3000

- 75x + 125y = 4200

-------------------------------

0 - 150y = - 1200

Divide both sides by - 150

-150y/-150 = -1200/-150

y = 8

Substitute y = 8 into the first equation

x + y = 42

x + 8 = 42

x = 42 - 8

x = 34

34 tickets were sold for 75 cents

8 tickets were sold for $1.25

3 0

2 years ago

An astronaut who weighs 174 lb on Earth weighs 29 lb on the moon. Which of the following are correct amounts for weight on Earth

Korolek [52]

174/29 = 6 times the difference

17*6 = 102

12*6= 72

34*6 =204

15*6 = 90

so the answer is A

5 0

3 years ago

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Sunspots have been observed for many centuries. Records of sunspots from ancient Persian and Chinese astronomers go back thousan

Lunna [17]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ $\leq$ 41

Alternate Hypothesis, $H_A$ : $\mu$ > 41

(b) The value of z test statistics is 1.08.

(c) We conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.

Step-by-step explanation:

We are given that in a random sample of 40 such periods from Spanish colonial times, the sample mean is x¯ = 47.0. Previous studies of sunspot activity during this period indicate that σ = 35.

It is thought that for thousands of years, the mean number of sunspots per 4-week period was about µ = 41.

Let $\mu$ = mean sunspot activity during the Spanish colonial period.

(a) Null Hypothesis, $H_0$ : $\mu$ $\leq$ 41 {means that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41}

Alternate Hypothesis, $H_A$ : $\mu$ > 41 {means that the mean sunspot activity during the Spanish colonial period was higher than 41}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean = 47

σ = population standard deviation = 35

n = sample of periods from Spanish colonial times = 40

So, the test statistics = $\frac{47-41}{\frac{35}{\sqrt{40} } }$

= 1.08

(b) The value of z test statistics is 1.08.

(c) Now, the P-value of the test statistics is given by;

P-value = P(Z > 1.08) = 1 - P(Z < 1.08)

= 1 - 0.8599 = 0.1401

Since, the P-value of the test statistics is higher than the level of significance as 0.1401 > 0.05, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.

8 0

3 years ago

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