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jeka57 [31]
3 years ago
6

Can someone please help me... I am totally confusedd!!!

Mathematics
1 answer:
Bas_tet [7]3 years ago
3 0
The length would be 6
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The beginning balance in a savings account us $64.98 after deposit of $73.87 what is the balance
Marina CMI [18]

the balance is $138.85 after the deposit

6 0
3 years ago
In △ABC , m∠A=53°,m∠B=17°, and a=27. Find the perimeter of the triangle.
sdas [7]
Using the Law of Sines  (sinA/a=sinB/b=sinC/c) and the fact that all triangles have a sum of 180° for their angles.

The third angle is C is 180-53-17=110°

27/sin53=b/sin17=c/sin110

b=27sin17/sin53, c=27sin110/sin53

And the perimeter is a+b+c so

p=27+27sin17/sin53+27sin110/sin53 units

p≈68.65 units  (to nearest hundredth of a unit)
6 0
3 years ago
The mean life of a television set is 119119 months with a standard deviation of 1414 months. If a sample of 7474 televisions is
Pepsi [2]

Answer:

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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 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 random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean

In this problem, we have that:

\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.6275

What is the probability that the sample mean would differ from the true mean by less than 1.11 months?

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

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

By the Central Limit Theorem

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

Z = \frac{120.1 - 119}{1.6275}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

X = 117.9

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

Z = \frac{117.9 - 119}{1.6275}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

3 0
3 years ago
HELP ME PLEASE
Mashutka [201]

Answer:

9 + \frac{n}{3} - 6 = 7

Step-by-step explanation:

Given

9 + \frac{n}{3} - 6

Required

Evaluate when n = 12

Substitute 12 for n in 9 + \frac{n}{3} - 6

9 + \frac{n}{3} - 6 = 9 + \frac{12}{3} - 6

Simplify the fraction

9 + \frac{n}{3} - 6 = 9 + 4 - 6

9 + \frac{n}{3} - 6 = 7

4 0
3 years ago
Find the sum of x^2+3x and <br> -2x^2+9x+5
Butoxors [25]

Answer: The answer is -x^2 + 12x + 5

7 0
3 years ago
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