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2 years ago

13

Suppose a temperature of −101°F was recorded in Antarctica and a temperature of 121°F was recorded in the Sahara Desert. How man

y degrees warmer is 121°F than −101°F?

1 answer:

lesya692 [45]2 years ago

7 0

222 degrees F, 121 minus -101 = 222

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Which of the following expressions is equal to 5^6/5^2? A) 5 • 5 • 5 • 5 B) 5 • 5 • 5 C) 1/5•5•5•5 D) 3

lana66690 [7]

Step-by-step explanation: To simplify, we will apply the <em>Quotient Rule</em>.

The 5's in this problem are bases so as you apply the quotient rule,

subtract the exponents but leave the base alone to get 5⁴.

We can also write 5⁴ as 5 · 5 · 5 · 5.

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3 years ago

Find the map ratio if 10 cm on the map is represented by 10 km

olga nikolaevna [1]

I think this is the answer :)

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3 years ago

Read 2 more answers

You have set a goal to drink more water every day. Your water bottle holds 0.5 liters of water. In the equation below, b is the

malfutka [58]

Answer:

Dependent Variable- l Independent Variable-b

Step-by-step explanation:

Dependent Variables are always Y and Independent Variables are always x.

In this case y=I and x=b. The amount of water your water bottle will hold will never change. But the amount of water you drink can change.

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3 years ago

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol

Julli [10]

Answer:

$0.071,1.928$

Step-by-step explanation:

Downtown Store North Mall Store

Sample size n 25 20

Sample mean $\bar{x}$ $9 $8

Sample standard deviation s $2 $1

$n_1=25\\n_2=20$

$\bar{x_1}=9\\ \bar{x_2}=8$

$s_1=2\\s_2=1$

$x_1-x_2=9-8=1$

Standard error of difference of means = $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Standard error of difference of means = $\sqrt{\frac{2^2}{25}+\frac{1^2}{20}}$

Standard error of difference of means = $0.458$

Degree of freedom = $\frac{\sqrt{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1}+\frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}$

Degree of freedom = $\frac{\sqrt{(\frac{2^2}{25}+\frac{1^2}{20}})^2}{\frac{(\frac{2^2}{25})^2}{25-1}+\frac{(\frac{1^2}{20})^2}{20-1}}$

Degree of freedom =36

So, z value at 95% confidence interval and 36 degree of freedom = 2.0280

Confidence interval = $(x_1-x_2)-z \times SE(x_1-x_2),(x_1-x_2)+z \times SE(x_1-x_2)$

Confidence interval = $1-(2.0280)\times 0.458,1+(2.0280)\times 0.458$

Confidence interval = $0.071,1.928$

Hence Option A is true

Confidence interval is $0.071,1.928$

4 0

3 years ago

Answer this for me please

dybincka [34]

Answer:

-3

Step-by-step explanation:

You cant count the spaces up (9) and over (-3) between the two points and do up/over so it would be 9/-3 and simplified that is -3/1 or 3

6 0

3 years ago