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Scrat [10]
3 years ago
11

Please help me with question no.1 and 2. chapter ; circles​

Mathematics
1 answer:
mars1129 [50]3 years ago
4 0
1. Find the perimeter of the square then the circumference of the circle then subtract the circle from the square.
Area of square=length of side squared=9.8^2=96.04.
Area of circle=(pi*radius)^2=((22/7)*(9.8/2))^2

2. Area of circle=pi*r squared
38.5=((22/7)*r)^2
Solve for r
Square root of 38.5/(22/7))=r
calculate area using above formula
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A set of data has the trend line modeled by the equation Y =1.5x + 11. What value of X corresponds to y =56?
erica [24]

Answer:

x = 30

Step-by-step explanation:

Plug in 56 in place of  y, then solve for x.

56 = 1.5x + 11

Subtract 11.

45 = 1.5x

Divide both sides by 1.5.

45/1.5 = x

30 = x

8 0
3 years ago
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Write 7/20 as a percent.<br><br> By what factor should you multiply the denominator and numerator?
levacccp [35]

Answer:

35%

Step-by-step explanation:

Use photomath it tells you everything.

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free 69 points A rectangular prism has a length of 8 meters, a width of 6 meters, and a height of 3 meters. Which equations coul
anyanavicka [17]

Answer:

V = 8 × 6 × 3

V = 48*3

Step-by-step explanation:

V = l*w*h

We know the length is 8, the width is 6 and the height is 3

V = 8*6*3

V = 48*3

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3 years ago
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Suppose a food scientist wants to determine whether two experimental preservatives result in different mean shelf lives for bana
melomori [17]

Answer:

The two sample t-test

Step-by-step explanation:

The appropriate test for thus is the two sample t test which is also known as the independent t test. This tests aims at determined whether there is a statistically significant difference between the means in two unrelated groups which in this context are a random sample with one type of preservative and another sample with another type of preservatives.

With this test, the researcher is able to compare the mean shelf lives of the bananas treated with the two different preservatives... The null hypothesis equalises the two means of the sample while the alternative does otherwise.

3 0
3 years ago
An auto company claims that the fuel efficiency of its sedan has been substantially improved. A consumer advocate organization w
zzz [600]

Answer:

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.983 -0}{\frac{1.685}{\sqrt{12}}}=2.021

df=n-1=12-1=11

p_v =P(t_{(11)}>2.021) =0.0342

If we compare the the p value with the significance level provided \alpha=0.1, we see that p_v < \alpha, so then we can reject the null hypothesis. and there is a significant increase in the miles per gallon from 2017 to 2019 at 10% of significance.

Step-by-step explanation:

Previous concepts

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

Let put some notation :

x=test value 2017 , y = test value 2019

x: 28.7 32.1 29.6 30.5 31.9 30.9 32.3 33.1 29.6 30.8 31.1 31.6

y: 31.1 32.4 31.3 33.5 31.7 32.0 31.8 29.9 31.0 32.8 32.7 33.8

Solution to the problem

The system of hypothesis for this case are:

Null hypothesis: \mu_y- \mu_x \leq 0

Alternative hypothesis: \mu_y -\mu_x >0

Because if we have an improvement we expect that the values for 2019 would be higher compared with the values for 2017

The first step is calculate the difference d_i=y_i-x_i and we obtain this:

d: 2.4, 0.3, 1.7,3,-0.2, 1.1, -0.5, -3.2, 1.4, 2, 1.6, 2.2

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{11.8}{12}=0.983

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =1.685

We assume that the true difference follows a normal distribution. The 4th step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.983 -0}{\frac{1.685}{\sqrt{12}}}=2.021

The next step is calculate the degrees of freedom given by:

df=n-1=12-1=11

Now we can calculate the p value, since we have a right tailed test the p value is given by:

p_v =P(t_{(11)}>2.021) =0.0342

If we compare the the p value with the significance level provided \alpha=0.1, we see that p_v < \alpha, so then we can reject the null hypothesis. and there is a significant increase in the miles per gallon from 2017 to 2019 at 10% of significance.

4 0
3 years ago
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