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puteri [66]
3 years ago
11

SOLVE

Mathematics
1 answer:
Sauron [17]3 years ago
7 0

Answer:

approximately

1623.6

Step-by-step explanation:

16 1/2%=.165 which is approximately equal to 1/6, so we can just multiply the original number by 6.

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Experimental data are collected as:
aliina [53]

Answer:

y = 1.00114x + 1.75243

Step-by-step explanation:

Given

The x and y values

Required

The regression line equation

Because of the length of the given data, I will run the analysis using online tools, then analyze the result.

From the analysis, we have:

\sum x = 5050

\sum y = 5231.1011

\bar x = 50.5

\bar y = 52.311

SSX = 83325 --- Sum of squares

SP = 83419.7626 --- Sum of products

The regression equation is calculated as:

y = ax + b

Where:

a = \frac{SP}{SSX}

So, we have:

a = \frac{83419.76}{83325}

a = 1.00114

b = \bar y - a * \bar x

b = 52.31 - (1.00114*50.5)

b = 1.75243

So:

y = ax + b becomes

y = 1.00114x + 1.75243

6 0
3 years ago
(25 points) Find the probability that a randomly selected student drives to school.
Gelneren [198K]

Answer:

2/5

Step-by-step explanation:

40 students drive to school

40/100 drive to school

2/5

6 0
3 years ago
Read 2 more answers
If an<br> angle has a measure of 43°, what is the<br> measure of its complement?
grigory [225]

Answer:

47

Step-by-step explanation:

6 0
3 years ago
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R
alexdok [17]
Thank you for the points
4 0
2 years ago
Daphne likes to ski at a resort that is open from December through April. According to a sign at the resort, 20, percent of the
Arturiano [62]

Answer:

Step-by-step explanation:

From the given information:

We can compute the  null hypothesis & the alternative hypothesis as:

{H_o}:\text{Distribution of snowfalls in her hometown is similar to claimed percentage }

{H_a}:\text{Distribution of snowfalls in her hometown is not similar to claimed percentage }

The degree of freedom = n - 1

The degree of freedom = 5 - 1

The degree of freedom = 4

At the level of significance of 0.05 and degree of freedom 4,

the rejection region = 9.488

However, we can compute the chi-square X² goodness of fit test as:

   

months  frequency (p)  observed O Expected E  Chi-square X^2= \dfrac{(O-E)^2}{E}

Dec          0.2                  16                   16                \dfrac{(16-16)^2}{16} =0      

Jan           0.250             11                   20                \dfrac{(11-20)^2}{20} =4.050      

Feb           0.200             16                  16                 \dfrac{(16-16)^2}{16} =0      

Mar           0.200             18                  16                 \dfrac{(18-16)^2}{16} =0.250

Apr           0.150               19                  12                 \dfrac{(19-12)^2}{12} =4.083

Total            1.000           80                 80                                  8.3833    

∴

The test statistics X² = 8.3833

Thus; we fail to reject the H_o since test statistics X² doesn't fall in the rejection region.

Therefore; there is sufficient evidence to conclude that the distribution of snowfalls in her hometown is not similar to the claimed percentage.

6 0
3 years ago
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