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Bond [772]
3 years ago
13

Plz help i give brainliest and points

Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
8 0

Answer:

b) 1 / 4

c) -3

Step-by-step explanation:

b) line passing through points (0,-1) (4,0)

slope = (y2 - y1) / (x2 - x1)

= (0 - (-1) ) / (4 - 0)

= 1 / 4

c) line passing through points (-1,2) (0, -1)

slope = (y2 - y1) / (x2 - x1)

= (-1 - 2) / (0 - (-1) )

= -3 / 1

= -3

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ludmilkaskok [199]
75% = 75/100÷ 5= 15/20 ÷5= ¾. does this help? hope it does
hope it helps :)
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3 years ago
Brenda needs to bake 12 cakes for a bake sale. Each cake requires 2 3/4 cups of flour. What is a reasonable estimate of the tota
marshall27 [118]
Brenda need ......... 24 36/48


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Suppose a batch of metal shafts produced in a manufacturing company have a population standard deviation of 1.3 and a mean diame
lbvjy [14]

Answer:

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

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 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.

In this problem, we have that:

\mu = 208, \sigma = 1.3, n = 60, s = \frac{1.3}{\sqrt{60}} = 0.1678

What is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

Lesser than 208 - 0.1 = 207.9 or greater than 208 + 0.1 = 208.1. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Lesser than 207.9.

pvalue of Z when X = 207.9. So

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

By the Central Limit Theorem

Z = \frac{207.9 - 208}{0.1678}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

2*0.2743 = 0.5486

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

6 0
3 years ago
4. A town has a population of 450,000 people. If 60% of the population are under 25 years of age, how many people in the town ar
lesya [120]

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2 years ago
Which interval is represented by the graph
kakasveta [241]

Answer:

you are correct the answer is c

4 0
3 years ago
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