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2 years ago

The value of a car decreases by 5% each year.

1 answer:

6 0

10,000*0.95^2
=9025

0.95 is a multiplyer since 100%-5%=95% then divided by 100 giving you 0.95

You always power by the amount of years

So the format is always
Amount*multiplyer^years

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A circle with radius 6 has a sector with a central angle of 48 what is the area of the sector

inn [45]

Answer:

≈ 15.08 units²

Step-by-step explanation:

The area (A) of the sector is calculated as

A = area of circle × fraction of circle

= πr² × $\frac{48}{360}$ ( r is the radius )

= π × 6² × $\frac{48}{360}$

= 36π × $\frac{48}{360}$ = π × $\frac{48}{10}$ = $\frac{48\pi }{10}$ ≈ 15.08

5 0

2 years ago

Read 2 more answers

In a recently administered iq test, the scores were distributed normally, with mean 100 and standard deviation 15. what proporti

iragen [17]

To solve for proportion we make use of the z statistic. The procedure is to solve for the value of the z score and then locate for the proportion using the standard distribution tables. The formula for z score is:

z = (X – μ) / σ

where X is the sample value, μ is the mean value and σ is the standard deviation

when X = 70

z1 = (70 – 100) / 15 = -2

Using the standard distribution tables, proportion is P1 = 0.0228

when X = 130

z2 = (130 – 100) /15 = 2

Using the standard distribution tables, proportion is P2 = 0.9772

Therefore the proportion between X of 70 and 130 is:

P (70<X<130) = P2 – P1

P (70<X<130) = 0.9772 - 0.0228

P (70<X<130) = 0.9544

Therefore 0.9544 or 95.44% of the test takers scored between 70 and 130.

6 0

3 years ago

Does anyone know this ??

alekssr [168]

Answer:

I believe that its C, but don't get mad at me if I get it wrong

8 0

1 year ago

Six friends attend a party. They form pairs for a game. How many different pairs are possible?

Leona [35]

Pairs, in this case, relates to a group of 2 or more. We have 6 friends. Let's call them A,B,C,D,E,F. This will allow us to make a [some sort of] combination tree:

1. ABC against DEF

2. ABD against CEF

3. ABE against CDF

4. ABF against CDE

5. ACD against BFE

6. ACE against BDF

7. ACF against BDE

8. ADE against BCF

9. ADF against BCE

10. AEF against BCD

I believe there are 12 combinations... I just can't think of the last 2 though.

5 0

2 years ago

What is the cubed root of 275÷7

almond37 [142]

Step-by-step explanation:

It's an irrational number.

$\sqrt[3]{275:7}=\sqrt[3]{\dfrac{275}{7}}=\dfrac{\sqrt[3]{275}}{\sqrt[3]{7}}=\dfrac{\sqrt[3]{275}\cdot\sqrt[3]{7^2}}{\sqrt[3]{7}\cdot\sqrt[3]{7^2}}=\dfrac{\sqrt[3]{275\cdot49}}{\sqrt[3]{7\cdot7^2}}\\\\=\dfrac{\sqrt[3]{13475}}{\sqrt[3]{7^3}}=\dfrac{\sqrt[3]{13475}}{7}=\dfrac{1}{7}\sqrt[3]{13475}$

3 0

2 years ago