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True [87]
3 years ago
6

What is the slope of the line through (-7,-6) and (-2,-5)

Mathematics
1 answer:
aliya0001 [1]3 years ago
6 0
Use the slope formula
\frac{ - 6 - ( - 5)}{ - 7 - ( - 2)}  \\   \frac{ - 6 + 7}{ - 7 + 2}  \\  \frac{1}{ - 5}
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2/5x+y=3<br> solve the equation for y.
stira [4]
2/5x+y=3

-2/5x        -2/5x

y=-2/5x+3

Does y = to whole number or just like that? if it Whole number I will solve it tell me
3 0
3 years ago
Consider the equation 1.5x + 4.5y = 18.
topjm [15]

Answer:

The slope is 0.3.

Step-by-step explanation:

1.5x + 4.5y = 18

<u>4.5y</u> = <u>18 - 1.5x</u>

4.5         4.5

y = 4 - 0.3x

3 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
3 years ago
Fill in the blank.
vlada-n [284]
C is your correct answer
4 0
3 years ago
Read 2 more answers
House prices in the neighborhood average at $90.05 per square foot. If the house has 1055 square feet, what should be its price?
telo118 [61]

Answer:

C. $95,000

Step-by-step explanation:

90.05 times 1055 is $95,000.75 the closest number to that is $95,000.

7 0
3 years ago
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