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3 years ago

in 25 minutes Li can run 10 laps around the track. consider the number of laps she can run per minute

Mathematics

2 answers:

Tresset [83]3 years ago

8 0

She can run 2.5 miles a minute

strojnjashka [21]3 years ago

7 0

Here you are going from 25 minutes to 1 minute; essentially, if you look at this algebraiclly you have 25 min=10 laps to 10 minutes= x laps; set up a proportion for this to get: 25 min/10 laps =1minutes/ x laps Solving by cross multiplying; multiply the denominator to the opposite numerator (this makes more sense when writing as a fraction; you will multiply in an "x" shape) then solve as the algebra problem of 25=10x ; divide both sides by 10 to isolate x and this will give you the answer of x=2.5

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Find the hypotenuse of each isosceles right triangle when the legs are of the given measure. 6 sqrt 2

Gemiola [76]

Isosceles right triangles have two equal sides (a and b) that are not the hypotenuse (c). And when two sides are equal, so are their opposite angles. There are only 180° degrees in any triangles, thus the right angle = 90°, so 90 left for the two equal, means that 2x=90,
x = 45°.

There are several ways to go about solving a triangle like this. The best and easiest is simply to memorize that the hypotenuse is exactly root2 times the other sides. Or, each isosceles side is the hypotenuse (c) ÷ root2
$a = b = c \div \sqrt{2} \\ c = a\sqrt{2} \\ c = 6 \sqrt{2} \times \sqrt{2} = 6 \times 2 = 12$
Another way to do it is the longer proof of Pythagorean Theorem:
${c}^{2} = {a}^{2} + {b}^{2}... \: \: c = \sqrt{({a}^{2} + {b}^{2})} \\$
$c= \sqrt{({6 \sqrt{2}) }^{2} + ({6 \sqrt{2})}^{2}} \\ = \sqrt{(2 \times{(6 \sqrt{2} )}^{2} )} = \sqrt{2(36 \times 2)} \\ c = \sqrt{144} = 12$

7 0

3 years ago

A number is chosen at random from 1-50. find the probability of selecting composite numbers

Ganezh [65]

.68 or a 68% chance

8 0

3 years ago

Read 2 more answers

For an outdoor concert by the city orchestra, concert organizers estimate that 16 comma 000 people will attend if it is not rai

Sladkaya [172]

Answer: 13300

========================================

Work Shown:

A = event that it rains

B = event that it does not rain

P(A) = 0.30

P(B) = 1-P(A) = 1-0.30 = 0.70

Multiply the attendance figures with their corresponding probabilities

if it rains, then 7000*P(A) = 7000*0.30 = 2100
if it doesn't rain, then 16000*P(B) = 16000*0.70 = 11200

Add up the results: 2100+11200 = 13300

This is the expected value. This is basically the average based on the probabilities. The average is more tilted toward the higher end of the spectrum (closer to 16000 than it is to 7000) because there is a higher chance that it does not rain.

3 0

3 years ago

Luda [366]

Answer:

a) H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

b) $\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

c) $\chi^2_{crit}=5.991$

d) Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

High school Some College Bachelor or higher Total

Public Broadcasting 15 15 10 40

Commercial stations 5 25 10 40

Total 20 40 20 80

We need to conduct a chi square test in order to check the following hypothesis:

Part a

H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

The level os significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

Part b

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{20*40}{80}=10$

$E_{2} =\frac{40*40}{80}=20$

$E_{3} =\frac{20*40}{80}=10$

$E_{4} =\frac{20*40}{80}=10$

$E_{5} =\frac{40*40}{80}=20$

$E_{6} =\frac{20*40}{80}=10$

And the expected values are given by:

High school Some College Bachelor or higher Total

Public Broadcasting 10 20 10 40

Commercial stations 10 10 20 40

Total 20 30 30 80

Part b

And now we can calculate the statistic:

$\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(2-1)(3-1)=2$

Part c

In order to find the critical value we need to look on the right tail of the chi square distribution with 2 degrees of freedom a value that accumulates 0.05 of the area. And this value is $\chi^2_{crit}=5.991$

Part d

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >33.75)=2.23x10^{-7}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(33.75,2,TRUE)"

Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

7 0

3 years ago

Mark jogged 3 5/7 km. His sister jogged 2 4/5 km. How much farther did Mark jog than his sister?

NeTakaya

Step-by-step explanation:

so therefore mark jogged 2 36/70km more than his sister

3 0

3 years ago