All categories

3 years ago

Please I need this asap :)

Mathematics

2 answers:

maksim [4K]3 years ago

7 0

It hour I had that question before and that the only right one

tiny-mole [99]3 years ago

3 0

I think its an hour i did the math fast in my head:)

You might be interested in

Regina has 24 inches of ribbon for a craft project. Tommy has 2 yards of ribbon. Choose the true statement below. (1 yard = 3 fe

seropon [69]

Answer:

Step-by-step explanation:

because 24 inches is 2 feet and 2 yards is is 6 feet

3 0

3 years ago

I need help ASAP on number 21

Nostrana [21]

Answer:

Step-by-step explanation:

It lines up with the intersection of the two lines

3 0

2 years ago

If a phone cost $224 what would it cost if there was a 7% sales tax.

Annette [7]

Answer:

$239.68

Step-by-step explanation:

Given

Cost = $224
Sales Tax = 7%

Solving

Cost w/ Tax = Cost (1 + Tax)
Cost w/ Tax = 224 (1 + 0.07)
Cost w/ Tax = 224 * 1.07
Cost w/ Tax = $239.68

5 0

2 years ago

Is the quotient 3.6 dividend 9 greater then or less than 1 exmplain to a friend

jekas [21]

Less than 1 because the numerator (3.6) is less than the denominator (9) and both are positive.

8 0

3 years ago

Question 1 options:Residents in Portland, Oregon think that their city has more rainfall than Seattle, Washington. To test this

dimaraw [331]

Answer:

There is no enough evidence to to support the claim that Portland has more average yearly rainfall than Seattle.

Being μ1: average rainfall in Portland, μ2: average rainfall in Seattle, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0$

P-value = 0.1290

As the P-value is bigger than the significance level, the effect is not significant and the null hypothesis failed to be rejected.

Step-by-step explanation:

We have to test the hypothesis of the difference between means.

The claim is that Portland has more average yearly rainfall than Seattle.

Being μ1: average rainfall in Portland, μ2: average rainfall in Seattle, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0$

The significance level is 0.10.

The sample for Portland, of size n1=45, has a mean of M1=37.50 and standard deviation of s1=1.82.

The sample for Seattle, of size n1=35, has a mean of M1=37.07 and standard deviation of s1=1.68.

The difference between means is:

$M_d= M_1-M_2=37.50-37.07=0.43$

The standard error for the difference between means is:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1.82^2}{45}+\dfrac{1.68^2}{35}}=\sqrt{ 0.0736+0.0688 }=\sqrt{0.1424}\\\\\\s_{M_d}=0.3774$

We can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.43-0}{0.3774}=1.1393$

The degrees of freedom are:

$df=n1+n2-2=45+35-2=78$

Then, the p-value for this one-tailed test with 78 degrees of freedom is:

$P-value=P(t>1.1393)=0.1290$

As the P-value is bigger than the significance level, the effect is not significant and the null hypothesis failed to be rejected.

There is no enough evidence to to support the claim that Portland has more average yearly rainfall than Seattle.

7 0

3 years ago