Express the ratio inits simplest form 1/3:1/2

zimovet [89]

It's 10 unless it's about something else

4 0

3 years ago

Lat year, 3,570 athlete finihed the Silvergrove Marathon. Thi year, the number of runner croing the finih line wa 20% higher. Ho

gizmo_the_mogwai [7]

The number of athletes that finished the race this year was 4284 athletes.

What is percentage?

10%, a comparative figure denoting one hundredth of any quantity. One percent, denoted by the symbol 1%, is equal to one hundredth of something; hence, 100 percent denotes the full thing, and 200 percent designates twice the amount specified.

Let x represent the amount of athletes that finished the race this year.

Last year, 3,570 athletes finished the Silvergrove Marathon. This year, the number of runners crossing the finish line was 20% higher. Hence:

x = 3570 + 20% of (3570)

x = 3570 + (0.2 * 3570)

x = 4284

4284 athletes finished the race.

To learn more about percentage from the given link

brainly.com/question/24877689

#SPJ4

6 0

1 year ago

Ayako and Joshua have a total of 59 sweets between them.

ehidna [41]

I'm pretty sure that the answer is 32.5 sweets is N.

4 0

3 years ago

What is 235x18 equal to

borishaifa [10]

Answer:

4230

Step-by-step explanation:

5 0

4 years ago

Read 2 more answers

Because of safety considerations, the FAA changed its guidelines for how small commuter airlines must estimate passenger weights

Alchen [17]

Answer:

$t=\frac{190-180}{\frac{23}{\sqrt{100}}}=4.348$

$df=n-1=100-1=99$

$p_v =2*P(t_{(99)}>4.348)=0.000033$

Since the p value obtained for this case is a very low value we have enough evidence to reject the null hypothesis that the true mean is equal to 180 at many of the possible significance levels commonly used. So then makes sense the claim that the true mean for the weigth is different from 180

Step-by-step explanation:

Information provided

$\bar X=190$ represent the mean weight

$s=23$ represent the sample standard deviation for the weight

$n=100$ sample size

$\mu_o =180$ represent the value to compare

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to determine if the true mean weight is different from 180 pounds, the system of hypothesis would be:

Null hypothesis: $\mu = 180$

Alternative hypothesis: $\mu \neq 180$

We don't know the population deviation for the variable of interest so then the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

Replacing the data given we got:

$t=\frac{190-180}{\frac{23}{\sqrt{100}}}=4.348$

Now we can find the p value but first we need to find the degrees of freedom given by:

$df=n-1=100-1=99$

Since we are conducting a two tailed test the p value can be calculated on this way:

$p_v =2*P(t_{(99)}>4.348)=0.000033$

Since the p value obtained for this case is a very low value we have enough evidence to reject the null hypothesis that the true mean is equal to 180 at many of the possible significance levels commonly used. So then makes sense the claim that the true mean for the weigth is different from 180

4 0

3 years ago