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EleoNora [17]
3 years ago
10

How do you write 9.51 x 10^-2 in standard form?

Mathematics
2 answers:
lbvjy [14]3 years ago
6 0
Move the decimal point 2 places to the left answer: 0.0951
Snezhnost [94]3 years ago
5 0

Answer:

.0951

Step-by-step explanation:

You move the decimal point 2 places to the left

9.51 --->.951--->.0951

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Consider a normal distribution curve where the middle 85 % of the area under the curve lies above the interval ( 8 , 14 ). Use t
NeTakaya

Answer:

\mu = 11

\sigma = 2.08

Step-by-step explanation:

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.

Middle 85%.

Values of X when Z has a pvalue of 0.5 - 0.85/2 = 0.075 to 0.5 + 0.85/2 = 0.925

Above the interval (8,14)

This means that when Z has a pvalue of 0.075, X = 8. So when Z = -1.44, X = 8. So

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

-1.44 = \frac{8 - \mu}{\sigma}

8 - \mu = -1.44\sigma

\mu = 8 + 1.44\sigma

Also, when X = 14, Z has a pvalue of 0.925, so when X = 8, Z = 1.44

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

1.44 = \frac{14 - \mu}{\sigma}

14 - \mu = 1.44\sigma

1.44\sigma = 14 - \mu

Replacing in the first equation

\mu = 8 + 1.44\sigma

\mu = 8 + 14 - \mu

2\mu = 22

\mu = \frac{22}{2}

\mu = 11

Standard deviation:

1.44\sigma = 14 - \mu

1.44\sigma = 14 - 11

\sigma = \frac{3}{1.44}

\sigma = 2.08

7 0
3 years ago
Which of the following inequalities has a solution x≥−3?
timama [110]
The 2nd one has the following inequality’s that has a solidity on x>-3
5 0
2 years ago
Majesty Video Production Inc. wants the mean length of its advertisements to be 26 seconds. Assume the distribution of ad length
Paladinen [302]

Answer:

a) By the Central Limit Theorem, approximately normally distributed, with mean 26 and standard error 0.44.

b) s = 0.44

c) 0.84% of the sample means will be greater than 27.05 seconds

d) 98.46% of the sample means will be greater than 25.05 seconds

e) 97.62% of the sample means will be greater than 25.05 but less than 27.05 seconds

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(also called standard error) 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 = 26, \sigma = 2, n = 21, s = \frac{2}{\sqrt{21}} = 0.44

a. What can we say about the shape of the distribution of the sample mean time?

By the Central Limit Theorem, approximately normally distributed, with mean 26 and standard error 0.44.

b. What is the standard error of the mean time? (Round your answer to 2 decimal places)

s = \frac{2}{\sqrt{21}} = 0.44

c. What percent of the sample means will be greater than 27.05 seconds?

This is 1 subtracted by the pvalue of Z when X = 27.05. So

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

By the Central Limit Theorem

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

Z = \frac{27.05 - 26}{0.44}

Z = 2.39

Z = 2.39 has a pvalue of 0.9916

1 - 0.9916 = 0.0084

0.84% of the sample means will be greater than 27.05 seconds

d. What percent of the sample means will be greater than 25.05 seconds?

This is 1 subtracted by the pvalue of Z when X = 25.05. So

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

Z = \frac{25.05 - 26}{0.44}

Z = -2.16

Z = -2.16 has a pvalue of 0.0154

1 - 0.0154 = 0.9846

98.46% of the sample means will be greater than 25.05 seconds

e. What percent of the sample means will be greater than 25.05 but less than 27.05 seconds?"

This is the pvalue of Z when X = 27.05 subtracted by the pvalue of Z when X = 25.05.

X = 27.05

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

Z = \frac{27.05 - 26}{0.44}

Z = 2.39

Z = 2.39 has a pvalue of 0.9916

X = 25.05

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

Z = \frac{25.05 - 26}{0.44}

Z = -2.16

Z = -2.16 has a pvalue of 0.0154

0.9916 - 0.0154 = 0.9762

97.62% of the sample means will be greater than 25.05 but less than 27.05 seconds

8 0
3 years ago
Triangle MNP is dilated according to the rule DO,1.5 (x,y)(1.5x, 1.5y) to create the image triangle M'N'P, which is not shown. W
scoray [572]

Answer: ?????????

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Find parametric equations and symmetric equations for the line. (Use the parameter t.) The line through (4, −5, 2) and parallel
Nataliya [291]

Answer:

Step-by-step explanation:

From the given information, the symmetric equations for the line pass through(4, -5, 2) i.e (x_o, y_o, z_o) and are parallel to \dfrac{x+5}{1} = \dfrac{y}{2}= \dfrac{z-3}{1}

The parallel vector to the line i + zj+k = ai + bj + ck

Hence, the equation for the line is :

x = x_o + at \\ \\ x = y_o + bt \\ \\ x = z_o + ct

x = 4 + t

y = -5 + 2t

z = 2 + t

Thus, x, y, z = ( 4+t, -5+2t, 2+t )

The symmetric equation can now be as follows:

\begin  {vmatrix} x = 4+ t   \\ \\  \dfrac{x-4}{1} = t  \begin {vmatirx} \end {vmatrix}\begin {vmatrix} y = - 5+2t  \\ \\ \dfrac{y+5}{2}  =t      \end {vmatrix}\begin {vmatrix} z =2+t  \\ \\ \dfrac{z-2}{1}  =t      \end {vmatrix}

∴

\dfrac{x-4}{1}= \dfrac{y+5}{2}=\dfrac{z-2}{1}

8 0
3 years ago
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