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Crazy boy [7]
3 years ago
8

28÷7+(-19+10)(-4) does anyone know how to do this ​

Mathematics
1 answer:
ikadub [295]3 years ago
5 0
Pemdas
28/7 + (-19+10)(-4)
28/7+ (-9)(-4)
28/7 + (36)
4+ (36)
40
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What’s 1500 times 150000
kotykmax [81]

Use long multiplication to evaluate.

225000000


4 0
3 years ago
Read 2 more answers
A consumer group has determined that the distribution of life spans for gas ovens has a mean of 15.0 years and a standard deviat
Mnenie [13.5K]

Answer:

B. Mean = 1.6 years, standard deviation = 0.92 years, shape: approximately Normal.

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Subtraction of normal variables:

When we subtract normal variables, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

35 gas ovens

A consumer group has determined that the distribution of life spans for gas ovens has a mean of 15.0 years and a standard deviation of 4.2 years. This means that:

\mu_G = 15, \sigma_G = 4.2, n = 35, s_G = \frac{4.2}{\sqrt{35}} = 0.71

40 electric ovens.

The distribution of life spans for electric ovens has a mean of 13.4 years and a standard deviation of 3.7 years.

\mu_E = 13.4, \sigma_E = 3.7, n = 40, s_E = \frac{3.7}{\sqrt{40}} = 0.585

Which of the following best describes the sampling distribution of barXG - bar XE, the difference in mean life span of gas and electric ovens?

By the Central Limit Theorem, the shape is approximately normal.

Mean: \mu = \mu_G - \mu_E = 15 - 13.4 = 1.6

Standard deviation:

s = \sqrt{s_G^2+s_E^2} = \sqrt{(0.71)^2+(0.585)^2} = 0.92

So the correct answer is given by option b.

3 0
3 years ago
Suri makes 15$ per hour and gets a weekly bonus of 25$.Juan makes 14$ per hour and gets weekly bonus of 50$.Is it possible for S
Setler [38]

Answer:

Yes, it is possible for Suri and Juan to make the same amount of wages if they both work 25 hours in one week.

Step-by-step explanation:

Let x = number of hours worked

Let y = total wages earned

Suri:  15x + 25 = y

Juan:  14x + 50 = y

Equate equations and solve for x:

15x + 25 = 14x + 50

Subtract 14x from both sides:  x + 25 = 50

Subtract 25 from both sides: x = 25

Yes, it is possible for Suri and Juan to make the same amount of wages if they both work 25 hours in one week.

3 0
3 years ago
Find the product.
rusak2 [61]
The product of (3p-7)(3p+7) is a. 9p^2-49
5 0
3 years ago
Read 2 more answers
Keith is working two jobs this summer as a lifeguard and a tutor. he earns $16 an hour as a lifeguard and $18 an hour as a tutor
Oxana [17]

Answer:

He could work 10 hours as a lifeguard, and 5 hours as a tutor

Step-by-step explanation:

If he makes $16 an hour as a lifeguard

            and $18 an hour as a tutor,

He could work 10 times $16, to make $160 for 10 hours.

Now we subtract this from $250, this makes $90.

We know need to find out how many times 18 goes into 90, 5 times.

So we know that he could work 10 hours as a lifeguard and 5 hours as a tutor.

I hope this is right.

I hope this helps too, have a nice day!

7 0
2 years ago
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