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jeyben [28]
3 years ago
9

Please help if you can and if you know how to help me understand these Im all ears.

Mathematics
2 answers:
maw [93]3 years ago
6 0

Aight let's answer this step by step

What you need to know is if two values are equal either one will be similar to the other it's only the format that's different if you calculate each of the fractions you'll see each pair is equal except for the last option

Where if calculated

8/3=2.67 but

5/4.8=1.04

Therefore values aren't equal to each other

LiRa [457]3 years ago
3 0
Values aren’t equal to each other
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Alex73 [517]

Answer:

the lesson will end at 11:10.

Step-by-step explanation:

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3 years ago
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In a study relating the degree of warping, in mm, of a copper plate (y) to temperature in °C (x), the following summary statisti
elena-14-01-66 [18.8K]

Answer:

a)

y = 0.3035 + 0.0082x

b)

0.6315 mm

c)

x = 23.9634 °C

Step-by-step explanation:

a. Compute the least-squares line for predicting warping from temperature. Round the answers to four decimal places.

We need to find an equation of the form

y = b + mx

where m is the slope and b the Y-intercept.

The slope m can be computed with the formula

\bf m=\displaystyle\frac{\displaystyle\sum_{i=1}^n(x_i-\bar x)(y_i-\bar y)}{\displaystyle\sum_{i=1}^n(x_i-\bar x)^2}

Replacing the values in our formula (we will round at the end of the calculations)

\bf m=\displaystyle\frac{806.94}{98,775}=0.008169476

the Y-intercept b is computed with the formula

\bf b=\bar y-m\bar x

therefore we have

\bf b=0.5188-0.008169476*26.36=0.303452613

and the least-squares line rounded to 4 decimals would be

y = 0.3035 + 0.0082x

b. Predict the warping at a temperature of 40°C. Round the answer to three decimal places.

We simply replace x with 40 to get

y = 0.3035 + 0.0082*40 = 0.6315 mm

c. At what temperature will we predict the warping to be 0.5 mm? Round the answer to two decimal places

Here, we replace y with 0.5 and solve for x

0.5 = 0.3035 + 0.0082x ===> x = (0.5-0.3035)/0.0082 ===>

 

x = 23.9634 °C

6 0
3 years ago
Previously, an organization reported that teenagers spent 24.5 hours per week, on average, on the phone. The organization thinks
Lena [83]

Answer:

We need to develop a one-tail t-student test ( test to the right )

We reject H₀  we find evidence that student spent more than 24,5 hours on the phone

Step-by-step explanation:

Sample size  n = 15     n < 30

And we were asked if the mean is higher than, therefore is a one-tail t-student test ( test to the right )

Population mean   μ₀  = 24,5

Sample mean   μ  =  25,7

Sample standard deviation s = 2

Hypothesis Test:

Null Hypothesis      H₀                             μ  =  μ₀

Alternative Hypothesis     Hₐ                  μ  >  μ₀

t (c) =  ?

We will define CI = 95 %  then   α = 5 %   α = 0,05    α/2 =  0,025

n = 15     then degree of freedom    df = 14

From t-student table  we get:  t(c) = 2,1448

And  t(s)

t(s) = ( μ  -  μ₀  ) / s/√n

t(s) = (25,7 - 24,5) /2/√15

t(s) = 2,3237

Now we compare   t(c)   and  t(s)

t(c)  =  2,1448         t(s)  = 2,3237

t(s) > t(c)

Then we are in the rejection region we reject H₀   we have evidence at 95% of CI that students spend more than 24,5 hours per week on the phone

8 0
3 years ago
12.03,1.2,12.3,1.203,12.301 order least to greatest
m_a_m_a [10]

Answer:

1,2, 1,203, 12,03, 12,3, 12,301

Step-by-step explanation:

1,2 → 1,200

1,203

12,3 → 12,300

12,301

I am joyous to assist you anytime.

8 0
3 years ago
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