You and your friend share a paper route you can deliver 4 papers in one minute

Sue's Sushi placed two orders with its fish supplier. One order was for 11 pounds of salmon and 6 pounds of tuna; the order tota

yawa3891 [41]

Divide the cost of everything by the amount of tuna and salmon. I don’t know the specific answer but that’s the main thing to do in this scenario would be to divide

8 0

3 years ago

•Use the Pythagorean Theorem c^6 = a2+ b2 •Show you work to find each missing side •

zhannawk [14.2K]

Answer:

hypotenuse = 10

Step-by-step explanation:

If you can tell at first glance it is a 3,4,5 triangle (6,8,10)/2=(3,4,5). Then you know that c has to equal 5.

If not,

a^2+b^2=c^2

6^2+8^2=c^2

36+64=c^2

100=c^2

10=c

6 0

3 years ago

on the highway Anita drove an average speed of 60.2 miles per hour at that speed about how far can she travel in three and a hal

Misha Larkins [42]

6. 64 hours so about 7 hours if you need to round up

hope this helps

please thank me

3 0

3 years ago

The magnitudes of the moon and sirius are -12.5 and -1.44, respectively. approximately how many times is the moon brighter than

Lelu [443]

Solution:

we are given that

The magnitudes of the moon and Sirius are -12.5 and -1.44 respectively.

we have been asked to find approximately how many times is the moon brighter than Sirius?

The word magnitude is used by the scientist to describe the brightness of moon/star/object etc.

So we can find the brightness by dividing the given values. so we get

$\frac{-12.5}{-1.44} =8.6805$

Hence we can say

The moon is 8.68 times brighter than Sirius.

3 0

3 years ago

Individuals filing federal income tax returns prior to March 31 had an average refund of $1102. Consider the population of "last

lions [1.4K]

Answer:

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

$H_0: \mu=1102\\\\H_a:\mu < 1102$

b) P-value = 0.0055

c) The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) Critical value tc=-1.96.

As t=-2.55, the null hypothesis is rejected.

Step-by-step explanation:

We have to perform a hypothesis test on the mean.

The claim is that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund ($1102).

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

$H_0: \mu=1102\\\\H_a:\mu < 1102$

b) The sample has a size n=600, with a sample refund of $1050 and a standard deviation of $500.

We can calculate the z-statistic as:

$t=\dfrac{\bar x-\mu}{s/\sqrt{n}}=\dfrac{1050-1102}{500/\sqrt{600}}=\dfrac{-52}{20.41}=-2.55$

The degrees of freedom are df=599

$df=n-1=600-1=599$

The P-value for this test statistic is:

$P-value=P(t$

c) Using a significance level α=0.05, the P-value is lower than the significance level, so the effect is significant. The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) If the significance level is α=0.025, the critical value for the test statistic is t=-1.96. If the test statistic is below t=-1.96, then the null hypothesis should be rejected.

This is the case, as the test statistic is t=-2.55 and falls in the rejection region.

4 0

3 years ago