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malfutka [58]
4 years ago
7

Model and solve 4÷2/3=

Mathematics
2 answers:
denis-greek [22]4 years ago
5 0
I hope this helps you



4/2/3


4×3/2


12/2


6
Phoenix [80]4 years ago
4 0
The answer to this question is 6

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2) Write the slope-intercept form of the equation of the line passing through the point (-5. -1) and perpendicular to
MAVERICK [17]
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3 years ago
Somebody Please help Me
Firdavs [7]
19. -x-2 = 2/3x + 3
5/3 x + 3 = -2
5/3x = -5
x = -3

20. none are correct, you can double check me by plugging in the x and y values in the coordanates into the first problem none of them worked out in the first equasion so no need to test the second

21. -3 is the answer, capable of being done by using desmos 
4 0
3 years ago
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
How many degrees has ABC been rotated counter-clockwise about the origin
natali 33 [55]
I feel like it’s 180 dregrees, good luck!
3 0
3 years ago
Read 2 more answers
The radii of two right circular cylinders are in the ratio 3 : 4 and their heights are in the ratio 1 : 2. Calculate the ratio o
bezimeni [28]

Answer:

18.852:50.272

Step-by-step explanation:

Step one:

given

The radii of two right circular cylinders are in the ratio 3 : 4

r1= 3

r2= 4

Their heights are in the ratio 1 : 2

h1= 1

h2= 2

Step two:

The expression for the curve surface area is

CSA= 2πrh

CAS1=  2πr1h1

CAS1=  2*3.142*3*1

CAS1=  18.852

CAS2=  2πr2h2

CAS2=  2*3.142*4*2

CAS1=  50.272

The ratio of their curved surface areas

=18.852:50.272

3 0
3 years ago
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