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MatroZZZ [7]
2 years ago
8

One afternoon in January, the temperature outside was -4*F. By midnight, the temperature had dropped by 10*F. what was the tempe

rature at midnight?
Brainliest
Mathematics
2 answers:
Yuliya22 [10]2 years ago
6 0

-4-10 = -14

The answer is: -14*F

Leokris [45]2 years ago
4 0

Answer:

i think the answer would be -14°F

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If a to the power x by y is equal to 1 then the value of x is​
Mazyrski [523]

Answer:

a^x/y=1              x: 0

Step-by-step explanation: w.k.t,        a^0=1( any variable raised to 0 is 1)

                                    so, here the exponent is x/y which should have been 0 so that answer was 1.

7 0
3 years ago
Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and
TiliK225 [7]

Answer:

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 117, \sigma = 10.6, n = 6, s = \frac{10.6}{\sqrt{6}} = 4.33

What is the level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L ?

This is the value of X when Z has a pvalue of 1-0.01 = 0.99. So X when Z = 2.33.

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

2.33 = \frac{X - 117}{4.33}

X - 117 = 2.33*4.33

X = 127.1

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

6 0
4 years ago
Formulating and solving inverse variation functions
andreyandreev [35.5K]

Answer:

See below.

Step-by-step explanation:

A. k = 50 * 3.6 = 180.

B. The equation is y = 180/x.

C. When x = 75, y = 180/75

= 2.4.

D.  Checking C:  2.4 = 180/75 = 2.4. Correct

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3 years ago
Which of the following describes the graph of {x | 2 < x < 2} ?
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Answer:

baahsuslskissisudjxnxnx

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2 years ago
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nydimaria [60]

Answer:

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Step-by-step explanation:

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Hope this helps you

4 0
3 years ago
Read 2 more answers
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