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3 years ago

-3b-(-2)=4b-12 show work please ASAP

2 answers:

3 0

Solve the equation:

– 3b – (– 2) = 4b <span>– 12

</span>– 3b + 2 = 4b – 12

– 3b = 4b – 12 – 2

– 3b = 4b – 14

– 3b – 4b = – 14

– 7b = – 14

– 14
b = ———
– 7

b = 2 <——— solution.

I hope this helps. =)

makkiz [27]3 years ago

3 0

-3(b) - (-2) - (4(b)) which is equal the number 0
-7(b) + 14
Now, lets solve → -7 = 0
b - 2 = 0
Next, we add the number 2 to each of the sides of your problem
Therefore, the result of b is 2

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1 year ago

In a one-month period, a convenience store

Vera_Pavlovna [14]

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3 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

NemiM [27]

Answer:

a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

Step-by-step explanation:

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 $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be 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.

Mildly obese

Normally distributed with mean 373 minutes and standard deviation 67 minutes. So $\mu = 373, \sigma = 67$

A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?

So $n = 5, s = \frac{67}{\sqrt{5}} = 29.96$

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

$Z = \frac{410 - 373}{29.96}$

$Z = 1.23$

$Z = 1.23$ has a pvalue of 0.8907.

So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

Lean

Normally distributed with mean 526 minutes and standard deviation 107 minutes. So $\mu = 526, \sigma = 107$

B) What is the probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes?

So $n = 5, s = \frac{107}{\sqrt{5}} = 47.86$

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

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$Z = -2.42$ has a pvalue of 0.0078.

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7 0

3 years ago

3) Chris wants to order DVD's over the internet. Each DVD costs $15.99 and shipping the entire order

satela [25.4K]

Answer:

He can buy 5 DVD's. (Brainliest please)

4 0

3 years ago

Read 2 more answers

Point C ∈ AB and AB=18 cm. The distance from C to A is 3 cm. shorter than the distance from C to B. Find AC and CB.

gogolik [260]

Answer:

AC=7.5 cm

CB=10.5cm

Step-by-step explanation:

Since we don't know the distance from C to B, we can label it as x. Our equation will be x+(x-3)=18. Simplify that so it is 2x-3=18. Add 3 to both sides and get 2x=21. x=10.5 So AC=7.5 cm and CB is 10.5 cm

7 0

3 years ago