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2 years ago

Need help with z-scores

Mathematics

2 answers:

zysi [14]2 years ago

6 0

hope this helps b.0.8937

ira [324]2 years ago

5 0

You need to look up the areas corresponding to the z-scores on a Standard Normal Distribution table. Using a table which gives areas to the LEFT of the z-score we work out the above shaded region area as follows

area = area to left for z score 0.85 - area to left for z score -1.2

= 0.8023 - 0.1151 = 0.6827

Its Option D.

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9.<br><br><br>How many times greater is the 6 in 6,275 than the 6 in 62.75?

joja [24]

Answer:100

Step-by-step explanation:

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1 year ago

Reca was approved for a credit card that charges an apr of 26% what is her monthly interest rate?

Lesechka [4]

Answer:

$i_m=0.01945$ or 1.945%

Step-by-step explanation:

Term annual percentage rate(APR) is the annual interest rate charged ona financial year for a duration of one year. APR can be converted to weekly, monthly, daily or even semi-annual rates using the below formula.

Effective rate for period = (1 + annual rate)(1 / n of periods) – 1

Rate is given as:

$i_m=(1+i)^\frac{1}{12}-1\\i_m=(1+0.26)^\frac{1}{12}-1\\i_m=0.01945$

7 0

2 years ago

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

7 0

1 year ago

The right rectangular prism below is made up of 8 cubes. Each cube has an edge length of 12 inch.

AysviL [449]

Answer:

1 in³

Step-by-step explanation:

Complete question

The right rectangular prism below is made up of 8 cubes each cube has an edge length of 1/2 inch what is the volume of this prism

Volume of a cube = l³

If each cube has an edge length of 1/2 inch, then:

Volume of each cube = 1/2³

Volume of each cube = 1/8

Volume of the prism = 8×1/8

Volume of the prism = 1 in³

4 0

2 years ago

I need the answers for this one as well!

Julli [10]

For the sake of showing work I will replace the empty space with an x like so...

$\frac{9}{10} =\frac{90}{x}$

To find out what x is you must cross multiply (aka butterfly)

***Image of this step is attached below

9*x = 10*90

9x = 900

Next divide 9 to both sides to finish isolating x. Since 9 is being multiplied by x, division (the opposite of multiplication) will cancel 9 out (in this case it will make 9 one) from the left side and bring it over to the right side.

9x ÷ 9 = 900 ÷ 9

1x = 100

x = 100

To make these fractions equivalent the empty spot must be 100

$\frac{9}{10} = \frac{90}{100}$

Hope this helped!

~Just a girl in love with Shawn Mendes

8 0

2 years ago