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1 year ago

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Suppose a distribution has a mean of 40, and a standard deviation of 5, if 3 is added to each value in the data what is the new

mean and standard divination?

1 answer:

Kamila [148]1 year ago

5 0

Answer:

Step-by-step explanation:

the mean would be 40+3= 43

the standard deviation wouldn't change

mean= 43

standard deviation= 5

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nikklg [1K]

Answer:

x = 30

Step-by-step explanation:

30 + 2x = 90

2x = 60

x = 30

8 0

2 years ago

WHAT THE ANSWER FOR THIS QUESTION? C = { X|X + 6 = 10}

Solnce55 [7]

Given that <span>C = { X | X + 6 = 10}

This can be interpreted grammatically as "C is a set X such that X is the solution to X + 6 = 10".

Given that X + 6 = 10
X = 10 - 6 = 4

Therefore, the solution is
C = {4}
</span>

7 0

3 years ago

A certain star is 2.4 x 102 light years from the Earth. One light year is about 5.9 x 1012 miles. How far

Studentka2010 [4]

Answer:

2.5 x $10^{15}$ miles

Step-by-step explanation:

We are given that 1 light year ≈ 5.9 x $10^{12}$ miles so:

4.3 x 10² light year · <u>5.9 x </u> $10^{12}$ <u> miles </u>
1 light year

Multiply 4.3 an 5.9 and round to 2 significant digits then add the exponents:

≈25 x $10^{14}$

This is not the scientific notation since the number is not between 0 and 10 so we move the decimal point by 1 place to the left. We then add 1 to the exponent to obtain the final answer:

≈ 2.5 x $10^{15}$ miles

6 0

2 years ago

Please help, thanks! Ok so, you pick a card at random, put it back, and then pick another card at random. There are FIVE cards a

tatyana61 [14]

Answer:

$P(x > 1\ and\ x < 2) = \frac{4}{25}$

Step-by-step explanation:

Given

$S = \{1,2,3,4,5\}$

$n(S) = 5$

Required

$P(x > 1\ and\ x < 2)$

$P(x > 1\ and\ x < 2)$ is calculated as:

$P(x > 1\ and\ x < 2) = P(x > 1) * P(x < 2)$

Since it is a probability with replacement, we have:

$P(x > 1\ and\ x < 2) = \frac{n(x > 1)}{n(S)} * \frac{n(x < 2)}{n(S)}$

For x > 1, we have:

$x > 1 = \{2,3,4,5\}\\$

$n(x > 1) = 4$

For x < 2, we have:

$x < 2 = \{1\}$

$n(x < 2) = 1$

$P(x > 1\ and\ x < 2) = \frac{n(x > 1)}{n(S)} * \frac{n(x < 2)}{n(S)}$

becomes

$P(x > 1\ and\ x < 2) = \frac{4}{5} * \frac{1}{5}$

$P(x > 1\ and\ x < 2) = \frac{4}{25}$

6 0

2 years ago

An amount of $23,000 is borrowed for 11 years at 6.75% interest, compounded annually. If the loan is paid in full at the end of

Pavel [41]

Answer:

Given that,

An amount of $23,000 is borrowed for 11 years at 6.75% interest, compounded annually.

To find the Amount paid back after 11 years.

Explanation:

we know that,

The formula to find the,

Amount after n years of interest rate r% compounded annually is,

$A=P(1+\frac{r}{100})^n$

where P is the initial amount.

Here, we have that,

P=$23,000

r=6.75

n=11

Substitute the values we get,

$A=23000(1+\frac{6.75}{100})^{11}$

Solving this we get,

$A=23000(1+0.0675)^{11}$ $A=23000(1.0675)^{11}$ $A=23000(2.05138)$ $A=47181.805$

Round to the nearest dollar.

$A=47181.805\approx47182$

Answer is: Amount paid back is $47182 after 11 years

7 0

1 year ago