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3 years ago

Karen deposits $450 into a savings account which earns simple interest at a rate of 5% per year. She wants to keep the

1 answer:

Taya2010 [7]3 years ago

7 0

450x.05=22.5
Multiply 22.5 by 5 years.
22.5x5=112.5

She’ll have $112.50 worth of interest earned after five years

B, C, and D

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A healthy human body is approximately 18% carbon by weight. Determine how many pounds of carbon a body weighing 200 lbs contains

dlinn [17]

Answer:

c. 36 lb. Of carbon; 16 kg of carbon

Step-by-step explanation:

An healthy body is approximately 18% carbon by weight. To determine how many pounds of carbon a body weighing 200Ibs contain can be gotten below.

A healthy person has approximately 18% carbon by weight. Therefore, the number of pounds of carbon for a person weighing 200 Ibs is

pounds of carbon = 18/100 × 200

pounds of carbon = 36 Ibs

According to the question the weight should be in kg.Therefore, we have to convert using the conversion rate you highlighted.

1 Ib = 0.454 kg

36 Ibs = ? kg

cross multiply

weight of carbon in kg = 0.454 × 36

weight of carbon in kg = 16.344 kg

weight of carbon in kg ≈ 16 kg

3 0

3 years ago

How do I graph (x−4)(x+2)

Anastasy [175]

(x - 4)(x + 2)

x^2 + 2x - 4x - 8

x^2 - 2x - 8

Let y = x^2 - 2x - 8

It is a parabola.

Take it from here.

6 0

2 years ago

List the probability value for each possibility in the binomial experiment calculated at the beginning of this lab, which was ca

Yanka [14]

Answer:

a. P(X = 0)= 0.001

b. P(X = 1)= 0.001

c. P(X=2)= 0.044

d. P(X=3)= 0.117

e. P(X=4)= 0.205

f. P(X=5)= 0.246

g. P(X=6)= 0.205

h. P(X=7)= 0.117

i. P(X=8)= 0.044

j. P(X=9)= 0.001

k. P(X=10)= 0.001

Step-by-step explanation:

Hello!

You have the variable X with binomial distribution, the probability of success is 0.5 and the sample size is n= 10 (I suppose)

If the probability of success p=0.5 then the probability of failure is q= 1 - p= 1 - 0.5 ⇒ q= 0.5

You are asked to calculate the probabilities for each observed value of the variable. In this case is a discrete variable with definition between 0 and 10.

You have two ways of solving this excersice

1) Using the formula

$P(X)= \frac{n!}{(n-X)!X!} * (p)^X * (q)^{n-X}$

2) Using a table of cummulative probabilities of the binomial distribution.

a. P(X = 0)

Formula:

$P(X=0)= \frac{10!}{(10-0)!0!} * (0.5)^0 * (0.5)^{10-0}$

P(X = 0) = 0.00097 ≅ 0.001

Using the table:

P(X = 0) = P(X ≤ 0) = 0.0010

b. P(X = 1)

Formula

$P(X=1)= \frac{10!}{(10-1)!1!} * (0.5)^1 * (0.5)^{10-1}$

P(X = 1) = 0.0097 ≅ 0.001

Using table:

P(X = 1) = P(X ≤ 1) - P(X ≤ 0) = 0.0107-0.0010= 0.0097 ≅ 0.001

c. P(X=2)

Formula

$P(X=2)= \frac{10!}{(10-2)!2!} * (0.5)^2 * (0.5)^{10-2}$

P(X = 2) = 0.0439 ≅ 0.044

Using table:

P(X = 2) = P(X ≤ 2) - P(X ≤ 1) = 0.0547 - 0.0107= 0.044

d. P(X = 3)

Formula

$P(X = 3)= \frac{10!}{(10-3)!3!} * (0.5)^3 * (0.5)^{10-3}$

P(X = 3)= 0.11718 ≅ 0.1172

Using table:

P(X = 3) = P(X ≤ 3) - P(X ≤ 2) = 0.1719 - 0.0547= 0.1172

e. P(X = 4)

Formula

$P(X = 4)= \frac{10!}{(10-4)!4!} * (0.5)^4 * (0.5)^{10-4}$

P(X = 4)= 0.2051

Using table:

P(X = 4) = P(X ≤ 4) - P(X ≤ 3) = 0.3770 - 0.1719= 0.2051

f. P(X = 5)

Formula

$P(X = 5)= \frac{10!}{(10-5)!5!} * (0.5)^5 * (0.5)^{10-5}$

P(X = 5)= 0.2461 ≅ 0.246

Using table:

P(X = 5) = P(X ≤ 5) - P(X ≤ 4) = 0.6230 - 0.3770= 0.246

g. P(X = 6)

Formula

$P(X = 6)= \frac{10!}{(10-6)!6!} * (0.5)^6 * (0.5)^{10-6}$

P(X = 6)= 0.2051

Using table:

P(X = 6) = P(X ≤ 6) - P(X ≤ 5) = 0.8281 - 0.6230 = 0.2051

h. P(X = 7)

Formula

$P(X = 7)= \frac{10!}{(10-7)!7!} * (0.5)^7 * (0.5)^{10-7}$

P(X = 7)= 0.11718 ≅ 0.1172

Using table:

P(X = 7) = P(X ≤ 7) - P(X ≤ 6) = 0.9453 - 0.8281= 0.1172

i. P(X = 8)

Formula

$P(X = 8)= \frac{10!}{(10-8)!8!} * (0.5)^8 * (0.5)^{10-8}$

P(X = 8)= 0.0437 ≅ 0.044

Using table:

P(X = 8) = P(X ≤ 8) - P(X ≤ 7) = 0.9893 - 0.9453= 0.044

j. P(X = 9)

Formula

$P(X = 9)= \frac{10!}{(10-9)!9!} * (0.5)^9 * (0.5)^{10-9}$

P(X = 9)=0.0097 ≅ 0.001

Using table:

P(X = 9) = P(X ≤ 9) - P(X ≤ 8) = 0.999 - 0.9893= 0.001

k. P(X = 10)

Formula

$P(X = 10)= \frac{10!}{(10-10)!10!} * (0.5)^{10} * (0.5)^{10-10}$

P(X = 10)= 0.00097 ≅ 0.001

Using table:

P(X = 10) = P(X ≤ 10) - P(X ≤ 9) = 1 - 0.9990= 0.001

Note: since 10 is the max number this variable can take, the cummulated probability until it is 1.

I hope it helps!

4 0

3 years ago

My father is 40, and my grandfather is 70. I am 2 years older than twice the gcf of my father's and my grandfather's ages. how o

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