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goldfiish [28.3K]
3 years ago
5

A student answered 86 problems on a test correctly and received a grade of 98%. How many problems were on the test, if all the p

roblems were worth the same number of points? (Round to the nearest whole number).
Mathematics
1 answer:
balu736 [363]3 years ago
7 0

Answer:

88 problems

Step-by-step explanation:

Set this problem up as fractions

86/x = 98/100

Cross multiply

86 × 100 = x × 98

8600 = 98x

Divide both sides by 98

87.75... = x

Round to nearest whole number

87.75 rounds up to 88

88 problems

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ethan made a rectangular prism that has a volume of 16 cubic inches. What is one set of possible dimensions for ethans rectangul
lions [1.4K]

Answer:

  0.5 inches by 0.5 inches by 64 inches

Step-by-step explanation:

Any three positive dimensions that have a product of 16 will be a set of possible dimensions. There are an infinite number of them. One such is listed above.

(If you have a list of answer choices, you can multiply them out to see which have a product of 16. We cannot guess what your choices might be.)

4 0
3 years ago
Otis and his 2 brother went to their favorite restaurant. Their total bill was $22.42. They used a coupon for 20% off of the tot
nevsk [136]

Answer:

$ 5.98

Step-by-step explanation:

20/100 × 22.42/1 = $ 4.48

$ 22.42 - $ 4.48 = $ 17.94

If they split that bill in two each would pay $ 5.98

I hope this helps.

4 0
3 years ago
Heights​ (cm) and weights​ (kg) are measured for 100 randomly selected adult​ males, and range from heights of 133 to 193 cm and
yawa3891 [41]

Answer:

The weight of an adult male who is 172 cm tall is 89.36 kg.

Step-by-step explanation:

The regression equation representing the relationship between height and weight of a person is:

\hat y=-105+1.13x

Here

<em>y</em> = weight of a person

<em>x</em> = height of a person

The information provided is:

\bar x=167.90\ cm\\\bar y=81.47\ kg\\r (X, Y) = 0.228\\p-value=0.023\\\alpha =0.05

The hypothesis to test the significance of the correlation between height and weight is:

<em>H₀</em>: There is no relationship between the height and weight, i.e. <em>ρ</em> = 0.

<em>Hₐ</em>: There is a relationship between the height and weight, i.e. <em>ρ </em>≠ 0.

Decision rule:

If the <em>p</em>-value of the test is less than the significance level, then the null hypothesis will be rejected and vice-versa.

According to information provided:

<em>p</em>-value = 0.023 < <em>α</em> = 0.05

The null hypothesis was rejected at 5% level of significance.

Thus, concluding that there is a relationship between the height and weight.

Compute the weight of an adult male with height, <em>x</em> = 172 cm as follows:

\hat y=-105+1.13x

  =-105+(1.13\times 172)\\=-105+194.36\\=89.36

Thus, the weight of an adult male who is 172 cm tall is 89.36 kg.

5 0
3 years ago
Can one of you lovely people help me please ?
saveliy_v [14]

Answer:

the answer is

C. = x=22.82

4 0
2 years ago
Read 2 more answers
Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.
adell [148]

Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

7 0
3 years ago
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