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Rashid [163]
3 years ago
7

A - g(x) = |x-5|+5 B - g(x) = |x-5| C - g(x) = |x|+5 D - g(x) = |x+5|+5

Mathematics
1 answer:
n200080 [17]3 years ago
5 0

Answer:

g(x) =  |x|  + 5

Step-by-step explanation:

Since both the original and transformed graph have the same x value for their vertex, there was no horizontal shift. However the vertex shifted up. Specifically it shifted up 5 units. Therefore:

g(x) = f(x) + 5

The plus 5 performs a translation up 5 units. Plugging in for f(x) it can be seen that:

g(x) =  |x|  + 5

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I report if u guess or answer randomly. Which confidence level would produce the widest interval when estimating the mean of a p
kap26 [50]

Answer:

B. 61%

Step-by-step explanation:

When we have the standard deviation of a sample, we use the t-distribution.

The width of an interval is related to the margin of error, which is given by the following equation:

M = T*s

In which T is related to the confidence level(and the sample size) and s is the standard deviation of the sample. Higher confidence levels have higher values of T.

A higher margin of error means that the interval is wider. To have a higher margin of error, we desire a higher value of T, which is achieved with a higher confidence level.

So the correct answer is:

B. 61%

5 0
2 years ago
Find the domain of the function f of x equals the square root of the quantity three x minus two.
katovenus [111]
F(x)=(3x-2)^0.5
C.x is greater than or equal to two thirds
8 0
2 years ago
Solve the system of the linear equation by adding or subtracting.
Natali [406]
X=1 and y=-2  

In ordered pairs it would be (1,0) and (0,-2)
5 0
3 years ago
The mean life of a television set is 119 months with a standard deviation of 14 months. If a sample of 74 televisions is randoml
irina [24]

Answer:

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 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, the sample means with size n of at least 30 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 = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.63

If a sample of 74 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 1.1 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}{1.63}

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}{1.63}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

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

8 0
2 years ago
100 BRAINLY POINTS
Troyanec [42]

Answer:

Given

f(x)=18,000(.88)^x

We need to find the value of x when f(x) <2000

2000 = 18000(0.88)^x

0.88^x 2000/18000

0.88^x 1/9

• x = log (1/9) / log (0.88)

• x= 17 (rounded down)

After 17 years or during year 18 the car value will drop below $2000

3 0
3 years ago
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