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ArbitrLikvidat [17]
3 years ago
7

What is the temperature change from 44 degrees Fahrenheit to -56 degrees Fahrenheit?

Mathematics
1 answer:
Vlad [161]3 years ago
8 0
The answer is 44-(-56)
44+55=99
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Please help! Solve x/12+1/3=−5/3. What is x?
Mama L [17]

the answer would be: x= -24

6 0
2 years ago
Read 2 more answers
PLEASE HELP ME ASAP <br><br> Determine the slope of each line:
Sladkaya [172]

Answer:

a=(2/5), b = (-2,6)

Step-by-step explanation:

well slope formula is rise/run. So for the first one you start at the point  rise 2 run 5 and you reach the endpoint. That is how you know you got the correct slope.  (2/5)

For the second one you start at the first point and you cannot go up because it is a negative slope you run 6 till you hit the other point go down 2. Since you go down 2 that would make it -2.  (-2,6)

5 0
3 years ago
Can i get some on these please?^^​
kotykmax [81]

Answer:

These should be right hopefully! Hope it helps!

Download pdf
6 0
2 years ago
Persons having Raynaud's syndrome are apt to suffer a sudden impairment of blood circulation in fingers and toes. In an experime
Citrus2011 [14]

Answer:

There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.

Step-by-step explanation:

We have to perform a hypothesis test on the difference between means.

The null and alternative hypothesis are:

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

μ1: mean heat output for subjects with the syndrome.

μ2: mean heat output for non-sufferers.

We will use a significance level of 0.05.

The difference between sample means is:

M_d=\bar x_1-\bar x_2=0.63-2.09=-1.46

The standard error is

s_{M_d}=\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}=\sqrt{0.3^2/9+0.5^2/9}=\sqrt{ 0.038 } \\\\ s_{M_d}=0.194

The t-statistic is

t=\dfrac{M_d}{s_{M_d}}=\dfrac{-1.46}{0.194}=-7.52

The degrees of freedom are

df=n_1+n_2-2=9+9-2=16

The critical value for a left tailed test at a significance level of 0.05 and 16 degrees of freedom is t=-1.746.

The t-statistic is below the critical value, so it lies in the rejection region.

The null hypothesis is rejected.

There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.

5 0
2 years ago
Help as soon as possible please
marta [7]

(3x + 5)° + (10x - 7)° = 180°

13x + (-2) = 180

13x = 180 + 2

13x = 182

x = 14

∠QRT = 3 * 14 + 5 = 42 + 5 = 47°

∠TRS = 10 * 14 - 7 = 140 - 7 = 133°

6 0
3 years ago
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