How can you write a constant rate of change?(Help I really don't get it )

1. Find a markdown of 1/9 on $450?

vovangra [49]

Step-by-step explanation:

1. 1/9 of 450

of is multiply

1/9 × 450

= ???

2. 15/60 = decimal number

change number to percentage = ???

I dont know what 3 is

put answers to 1 and 2 in comments i will mark if u don't understand comment

please mark brainiest

5 0

2 years ago

Explain when it might be more efficient to use the percent equation rather than the percent proportion.

Tamiku [17]

In percent equation method , we transformed the number that we wanted to look for the percentage of into decimals and simply move the commas into two number on the right
If you get a hang on it, this method will be faster than the percent equation

hope this helps

4 0

3 years ago

If your chances of winning a game are 4 out of 10, that’s better than having chances of 2 out of 10.

Nina [5.8K]

Answer:

It is true.

Step-by-step explanation:

4/10 which is 0.4 is a higher probability than 2/10 which is 0.2.

Hope it helped :)

6 0

3 years ago

If the circumference of a circle is 120 inches and M angle B equals 18° then what is the length of arc AC

Andru [333]

Answer:

For a circle of radius R, the circumference is:

C = 2*pi*R

where pi = 3.14

And if we have an arc defined by an angle θ, the length of the arc is:

A = (θ/360°)*2*pi*R

Here we can not see the image, then i assume that B is the angle that defines the arc AC.

Now we know that the circumference is 120 in, then:

2*pi*R = 120in

Then the length of the arc is:

A = (θ/360°)*120 in

Then if the angle is 18°, we have:

A = (18°/360)*120 in = 6in

6 0

3 years ago

Congress regulates corporate fuel economy and sets an annual gas mileage for cars. A company with a large fleet of cars hopes to

attashe74 [19]

Answer:

a) Null hypothesis: $\mu \leq 30.2$

Alternative hypothesis: $\mu > 30.2$

b) $X \sim N(\mu=32.12, \sigma=4.83)$

And the distribution for the random sample is given by:

$\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)$

c) $t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81$

$p_v =P(t_{(49)}>2.81)=0.0035$

d) If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.

e) The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

$\bar X=32.12$ represent the sample mean

$s=4.83$ represent the sample standard deviation

$n=50$ sample size

$\mu_o =30.2$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

a. Define the parameter and state the hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 30.2 mpg, the system of hypothesis would be:

Null hypothesis: $\mu \leq 30.2$

Alternative hypothesis: $\mu > 30.2$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

b. Define the sampling distribution (mean and standard deviation).

Let X the random variable who represent the variable of interest. And we know that the distribution for X is:

$X \sim N(\mu=32.12, \sigma=4.83)$

And the distribution for the random sample is given by:

$\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)$

c. Perform the test and calculate P-value

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=50-1=49$

Since is a one right tailed test the p value would be:

$p_v =P(t_{(49)}>2.81)=0.0035$

d. State your conclusion.

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.

e. Explain what the p-value means in this context.

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.

6 0

3 years ago