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It's -25 in Alaska and 75 in Florida. How much hotter is it in Florida than in Alaska?

emmasim [6.3K]

Just subtract to find the difference.

75 - (-25) = 75 + 25 = 100.

It is 100 degrees hotter in florida than alaska.

3 0

3 years ago

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There are 15 hats on a rack and 9 of them are orange<br>What percentage of the hats are NOT orange?

dmitriy555 [2]

Answer:

40%

Step-by-step explanation:

15-9=6

6/15=2/5=40/100=40%

4 0

3 years ago

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Anyone? will mark you as brainlist

scoray [572]

Answer:

true

Step-by-step explanation:

5 0

3 years ago

For each function, find f(-1), f(1), and f(3)<br> f(x)=3x-4

Anna [14]

Answer:

f(x)= 3x-4

f(-1)= 3*-1-4

f(-1)= 2-4

f(-1)= -2

f(x)= 3x-4

f(3)= 3*3-4

f(3)= 9-4

f(3)=5

f(x)= 3x-4

f(1)= 3*1-4

f(1)= 3-4

f(1)= -1

Step-by-step explanation:

6 0

3 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

3 years ago