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Mumz [18]
3 years ago
14

What is 7(-3). I am confuse and don't understand.

Mathematics
1 answer:
emmasim [6.3K]3 years ago
7 0
7(-3) is an example of multiplication. What the question wants you to do is multiply the outside number (7) by the inside number (-3).
Basically you are just multiplying -3 x 7.
Your product is -21
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ZA and B are supplementary
ANEK [815]

Answer:

< A+< B =180

2n + 3n +20 =180

5n = 160

n = 32

(B= 116 ..........

answer is C

5 0
3 years ago
Complete the equation so that it has no solution 9(x-4)-5x=blank-30
Troyanec [42]

96.1 i dont know if its right but try it

4 0
2 years ago
Solve 10 X + 8 / 2 X + 7 is equal to 4 ​
ale4655 [162]

Answer:

x = 10

Step-by-step explanation:

\frac{10x + 8}{2x + 7}  = 4

or, 10x + 8 = 4(2x + 7)

or, 10x + 8 = 8x + 28

or, 10x - 8x = 28 -8

or, 2x = 20

or, x = 20/2

or, x = 10

x = 10

4 0
2 years ago
Is (3,6) a solution of y's 3x + 4?
notka56 [123]

Step 1:

(3,6) means x coordinate is 3 and y coordinate 6

Step 2:

We substitute the values of x = 3 and y = 6 into the inequality

\begin{gathered} y\text{ }\leq\text{ }\frac{2}{3}\text{ }x\text{ + 4} \\ 6\text{ }\leq\text{ }\frac{2}{3}\text{ x }\frac{3}{1}\text{  + 4} \\ 6\text{ }\leq\text{ 2 + 4} \\ 6\text{ }\leq\text{ 6} \end{gathered}

Conclusion:

Since 6 is equal to 6, then (3,6) is a solution of the given Inequality

3 0
1 year 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
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