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Diano4ka-milaya [45]

3 years ago

11

The scores on a mathematics exam have a mean of 70 and a stadard deviation of 7. Find the x-value that corresponds to the z-scor

e -1.45.

1 answer:

mylen [45]3 years ago

7 0

Answer:

59.85

Step-by-step explanation:

The scores on a mathematics exam have a mean of 70 and a stadard deviation of 7. Find the x-value that corresponds to the z-score -1.45.

We solve the above question using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean = 70

σ is the population standard deviation = 7

z = -1.45

Hence:

-1.45 = x - 70/7

Cross Multiply

-1.45 × 7 = x - 70

-10.15 = x - 70

x = -10.15 + 70

x = 59.85

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A business uses straight-line depreciation to determine the value y of an automobile over a 5-year period. Suppose the original

Elena-2011 [213]

For this problem. it is found that:

a) The automobile has depreciated 75% in 5 years.

b) The value of the automobile was reduced by $10,500 in five years.

c) The linear function is: s(t) = 14,500 - 2,100t.

<h3>What is a linear function?</h3>

A linear function is modeled by:

y = ax + b

In which:

a is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.

b is the y-intercept, which is the value of y when x = 0, and can also be interpreted as the initial value of the function.

From the problem, we have that:

a) The automobile has depreciated 75% in 5 years, as 4000/14500 = 0.25, and 100% - 25% = 75%.

b) The value of the automobile was reduced by $10,500 in five years, as 14500 - 4000 = $10,500.

For the linear function, we have that:

The initial value is b = 14500.

The slope is of a = -10500/5 = -2100.

Hence the linear function is: s(t) = 14,500 - 2,100t.

More can be learned about linear functions at brainly.com/question/24808124

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6 0

2 years ago

Find the differences between the values and the mean. then square each difference 39 42 45 47 52

Karolina [17]

The correct answer is 45

8 0

4 years ago

What would the number 8671.42578125 ten be in IEEE 754 single precision floating point format. You need to follow the following

Arlecino [84]

Answer:

Step-by-step explanation:

Answers are explained/Solved in the attach document.

5 0

4 years ago

Evaluate the integral ∫2032x2+4dx. Your answer should be in the form kπ, where k is an integer. What is the value of k? (Hint: d

faltersainse [42]

Here is the correct computation of the question;

Evaluate the integral :

$\int\limits^2_0 \ \dfrac{32}{x^2 +4} \ dx$

Your answer should be in the form kπ, where k is an integer. What is the value of k?

(Hint: $\dfrac{d \ arc \ tan (x)}{dx} =\dfrac{1}{x^2 + 1}$ )

k = 4

(b) Now, lets evaluate the same integral using power series.

$f(x) = \dfrac{32}{x^2 +4}$

Then, integrate it from 0 to 2, and call it S. S should be an infinite series

What are the first few terms of S?

Answer:

(a) The value of k = 4

(b)

$a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}$

Step-by-step explanation:

(a)

$\int\limits^2_0 \dfrac{32}{x^2 + 4} \ dx$

$= 32 \int\limits^2_0 \dfrac{1}{x+4}\ dx$

$=32 (\dfrac{1}{2} \ arctan (\dfrac{x}{2}))^2__0$

$= 32 ( \dfrac{1}{2} arctan (\dfrac{2}{2})- \dfrac{1}{2} arctan (\dfrac{0}{2}))$

$= 32 ( \dfrac{1}{2}arctan (1) - \dfrac{1}{2} arctan (0))$

$= 32 ( \dfrac{1}{2}(\dfrac{\pi}{4})- \dfrac{1}{2}(0))$

$= 32 (\dfrac{\pi}{8}-0)$

$= 32 ( (\dfrac{\pi}{8}))$

$= 4 \pi$

The value of k = 4

(b) $\dfrac{32}{x^2+4}= 8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -... \ \ \ \ \ (Taylor\ \ Series)$

$\int\limits^2_0 \dfrac{32}{x^2+4}= \int\limits^2_0 (8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -...) dx$

$S = 8 \int\limits^2_0dx - \dfrac{3}{2^1} \int\limits^2_0 x^2 dx + \dfrac{3}{2^3}\int\limits^2_0 x^4 dx - \dfrac{3}{2^5}\int\limits^2_0 x^6 dx+ \dfrac{3}{2^7}\int\limits^2_0 x^8 dx-...$

$S = 8(x)^2_0 - \dfrac{3}{2^1*3}(x^3)^2_0 +\dfrac{3}{2^3*5}(x^5)^2_0- \dfrac{3}{2^5*7}(x^7)^2_0+ \dfrac{3}{2^7*9}(x^9)^2_0-...$

$S= 8(2-0)-\dfrac{1}{2^1}(2^3-0^3)+\dfrac{3}{2^3*5}(2^5-0^5)- \dfrac{3}{2^5*7}(2^7-0^7)+\dfrac{3}{2^7*9}(2^9-0^9)-...$

$S= 8(2-0)-\dfrac{1}{2^1}(2^3)+\dfrac{3}{2^3*5}(2^5)- \dfrac{3}{2^5*7}(2^7)+\dfrac{3}{2^7*9}(2^9)-...$

$S = 16-2^2+\dfrac{3}{5}(2^2) -\dfrac{3}{7}(2^2) + \dfrac{3}{9}(2^2) -...$

$S = 16-4 + \dfrac{12}{5}- \dfrac{12}{7}+ \dfrac{12}{9}-...$

$a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}$

6 0

3 years ago

-7x - 2y = 14<br><br> 6x + 6y = 18<br><br> x =<br><br> y =

Mrac [35]

Answer:

6(2) + 6(1) = 18

Step-by-step explanation:

6 0

4 years ago