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3 years ago

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

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If an<br> angle has a measure of 43°, what is the<br> measure of its complement?

grigory [225]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

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