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alexandr1967 [171]
3 years ago
11

In the diagram above

Mathematics
2 answers:
Inessa [10]3 years ago
8 0

Answer:

140

Step-by-step explanation:

First you have to solve for x

20x + 60 = 30x + 20

so you would subtract 20x from both sides

60 = 10x + 20

then subtract 20 from both sides

40 = 10x

then divide both sides by 10

4 = x

then you would put 4 in for x in the equation for angle 1

20(4) + 60

80 + 60

140

Xelga [282]3 years ago
6 0

Answer:

angle 1=20(4)+60=140 degrees

Step-by-step explanation:

theory : If two parallel lines are cut by a transversal, the corresponding angles are congruent.

so angle 1 and angle 3 are equal:

20x+60=30x+20

60-20=30x-20x

40=10x

x=40/10=4

angle 1 = 20(x)+60

angle 1=20(4)+60=140 degrees

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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
It takes Mr. Martin 12 minutes to drive his bus route without stopping to pick up any passengers. For each passenger stop, he es
PtichkaEL [24]
14 = 0.5p + 12 <== ur equation
8 0
3 years ago
Read 2 more answers
Solve the equation for x.
DiKsa [7]
You multiply 0.6 by 5x and 10 and you get 3x+6=-24. you then subtract 6 from both sides of the equation and you get 3x=-30. after that you divide both sides by 3 and you get x=-10. so the answer is B!
6 0
3 years ago
What is the answer x+5&gt;23
olga2289 [7]
18<x is the answer you are looking for. Glad to help.
3 0
3 years ago
A car company is going to issue new ID codes to its employees. Each code will have one digit followed by four letters. The lette
scoray [572]

2734375 employee ID codes can be generated

<em><u>Solution:</u></em>

Given that Each code will have one digit followed by four letters

The letter V and the digits 3, 4, and 5 will not be used

So, there are 25 letters and 7 digits that will be used

Assume that the letters can be repeated

Each code = 1 digit + four letters

digits 3 , 4 , 5 cannot be used

Therefore, the remaining digits are 1, 2, 6, 7, 8 , 9, 0

7 digits that will be used

One digit can be chosen from 7 digits in 7 ways

The first letter can be chosen from 25 letters in 25 ways

Since given that letters can be repeated

The second letter can be chosen from 25 letters in 25 ways

Similarly, the third letter can be chosen from 25 letters in 25 ways

Similarly, the fourth letter can be chosen from 25 letters in 25 ways

<u><em>Therefore, the total number of ways are:</em></u>

\rightarrow 7 \times 25 \times 25 \times 25 \times 25 = 7 \times 25^4\\\\ \rightarrow 2734375

Thus 2734375 employee ID codes can be generated

3 0
3 years ago
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