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Reika [66]
3 years ago
11

The high temperature on Monday in Alaska was -2 degrees. Tuesday's high temperature was 3 degrees lower than Monday's. What was

Tuesday's high temperature?
A) 5 degrees
B) -5 degrees
C) 1 degree
D) 1 degree below zero
Mathematics
1 answer:
oee [108]3 years ago
3 0

Answer:

The answer is B) -5

Step-by-step explanation:

It is -5 because if you take monday's temperature and subtract it by tuesday's temperature. You get -5

-2-3=-5

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For the given hypothesis test, determine the probability of a Type II error or the power, as specified. A hypothesis test is to
erica [24]

Answer:

the probability of a Type II error if in fact the mean waiting time u, is 9.8 minutes is 0.1251

Option A) is the correct answer.

Step-by-step explanation:

Given the data in the question;

we know that a type 11 error occur when a null hypothesis is false and we fail to reject it.

as in it in the question;

obtained mean is 9.8 which is obviously not equal to 8.3

But still we fail to reject the null hypothesis says mean is 8.3

Hence we have to find the probability of type 11 error

given that; it is right tailed and o.5, it corresponds to 1.645

so

z is equal to 1.645

z = (x-μ)/\frac{S}{\sqrt{n} }

where our standard deviation s = 3.8

sample size n = 50

mean μ = 8.3

we substitute

1.645 = (x - 8.3)/\frac{3.8}{\sqrt{50} }

1.645 = (x - 8.3) / 0.5374

0.884023 = x - 8.3

x = 0.884023 + 8.3

x = 9.18402

so, by general rule we will fail to reject the null hypothesis when we will get the z value less than 1.645

As we reject the null hypothesis for right tailed test when the obtained test statistics is greater than the critical value

so, we will fail to reject the null hypothesis as long as we get the sample mean less than 9.18402

Now, for mean 9.8 and standard deviation 3.8 and sample size 50

Z =  (9.18402 - 9.8)/\frac{3.8}{\sqrt{50} }

Z = -0.61598 / 0.5374

Z = - 1.1462 ≈ - 1.15

from the z-score table;

P(z<-1.15) = 0.1251

Therefore, the probability of a Type II error if in fact the mean waiting time u, is 9.8 minutes is 0.1251

Option A) is the correct answer.

8 0
3 years ago
At 106°F, a certain insect chirps at a rate of 67 times per minute, and at 108°F, they chirp 83 times per minute. Write an equat
Step2247 [10]

Answer:

y=8x-781

Here, x represents temperature and y denotes rate of chirping per minute.

Step-by-step explanation:

Let x represents temperature and y denotes rate of chirping per minute.

At 106°F, a certain insect chirps at a rate of 67 times per minute.

Take (x_1,y_1)=(106,67)

At 108°F, they chirp 83 times per minute.

Take (x_2,y_2)=(108,83)

Slope intercept form:

y-y_1=(\frac{y_2-y_1}{x_2-x_1})(x-x_1)

y-67=(\frac{83-67}{108-106})(x-106)\\\\y-67=(\frac{83-67}{108-106})(x-106)\\\\y-67=(\frac{16}{2})(x-106)\\\\y-67=8(x-106)\\y-67=8x-848\\y=8x+67-848\\y=8x-781

3 0
3 years ago
A fulcrum moving a resistance of 200 g has a distance to the fulcrum of 20 cm, the effort mass of 50 g has a distance to the ful
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Answer:

The ideal mechanical advantage (IMA) is 4.

Step-by-step explanation:

The ideal mechanical advantage is the ratio of length of longer lever L_e to that of shorter lever L_r

IMA \frac{L_e}{L_r}

Please refer to the image attached.

We could see that the the resistance load moves 10\ cm cm towards the fulcrum so the distance of resistance load from fulcrum = (20-10) =10\ cm

Now the as the effort force moves 40\ cm towards the fulcrum overall distance from the fulcrum to the effort force (load) =(80-40)=40\ cm

Plugging the values of the distances in IMA formula we can have.

IMA =\frac{(80-40)}{(20-10)} =\frac{40}{10}  =4.

So the IMA of the fulcrum (simple machine) = 4

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4 years ago
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The answer is 0.97

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