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Flauer [41]
4 years ago
13

Solve. –30 = 6z a. –5 b. –6 c. –24 d. –180

Mathematics
2 answers:
Sergeu [11.5K]4 years ago
8 0
-30 = 6z

Plug in 6(-5) and you get -30

Answer is A
masya89 [10]4 years ago
5 0
The correct answer is A. -5

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Answer: probably 4 cards each person

Step-by-step explanation:

8 0
3 years ago
Find the value of x. Diagram not to scale
lara31 [8.8K]
We first need to find the sum of the interior angles in a pentagon, which you do by doing (5 - 2) × 180, which equals 3 × 180 = 540°. Now you know that all of the angles inside the shape must add up to 540, so you can make an equation:
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8 0
3 years ago
Read 2 more answers
A 40% sugar solution is added to an 85% sugar solution to create 1800 mL of a 60% solution. How much of each solution is used? (
denis-greek [22]

Answer:

Number of ml of 40% sugar = x = 1000mL

Number of ml of 85% sugar used = y = 800mL

Step-by-step explanation:

Let the

Number of ml of 40% sugar = x

Number of ml of 85% sugar used = y

From the above question, our system of equations is given as:

x + y = 1800mL ....... Equation 1

x = 1800 - y

40% × x + 85% × y = 60% × 1800mL

0.4x + 0.85y = 1080.... Equation 2

We substitute 1800 - y for x in Equation 2

0.4(1800 - y) + 0.85y = 1080

720 - 0.4y + 0.85y = 1080

- 0.4y + 0.85y = 1080 - 720

0.45y = 360

y = 360/0.45

y = 800mL

Solving for x

x = 1800 - y

x = 1800 - 800

x = 1000mL

Therefore,

Number of ml of 40% sugar = x = 1000mL

Number of ml of 85% sugar used = y = 800mL

4 0
3 years ago
During the past six months, 73.2 percent of US households purchased sugar. Assume that these expenditures are approximately norm
lina2011 [118]

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Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

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 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 s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

4 0
4 years ago
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