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1 year ago

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Olivia deposits $6,952 in a savings account paying 3.72% interest. To the nearest dollar, how much money does Olivia have in tot

al after sixteen years?a.$4,138b.$8,568c.$11,090d.$29,901

2 answers:

liubo4ka [24]1 year ago

6 0

Answer:

C on EDGE 2021

Step-by-step explanation:

ivolga24 [154]1 year ago

4 0

Answer:

11090

Step-by-step explanation:

6952x.0372=258.61

258.61x16=4137.83

4137.83+6952=11090

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On average, a pair of bike tires can last 18 months if used every day. The length of time bike tires can last is exponentially d

aleksandr82 [10.1K]

Answer:

We conclude that eighty percent of bike tires last at most 14.4 months.

Step-by-step explanation:

We know that a pair of bike tires can last 18 months if used every day.

We calculate how much eighty percent of the tire will last in months.

We have the following proportion:

$18:100\%=x:80\%\\\\100x=18\cdot 80\\\\100x=1440\\\\x=\frac{1440}{100}\\\\x=14.4$

We conclude that eighty percent of bike tires last at most 14.4 months.

5 0

2 years ago

5 1 - 2 5 4 7= please help asap

Yuki888 [10]

Answer:

-2496

Step-by-step explanation:

6 0

2 years ago

Probabilities with possible states of nature: s1, s2, and s3. Suppose that you are given a decision situation with three possibl

amm1812

Answer:

1. $P(s_1|I)=\frac{1}{11}$

2. $P(s_2|I)=\frac{8}{11}$

3. $P(s_3|I)=\frac{2}{11}$

Step-by-step explanation:

Given information:

$P(s_1)=0.1, P(s_2)=0.6, P(s_3)=0.3$

$P(I|s_1)=0.15,P(I|s_2)=0.2,P(I|s_3)=0.1$

(1)

We need to find the value of P(s₁|I).

$P(s_1|I)=\frac{P(I|s_1)P(s_1)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_1|I)=\frac{(0.15)(0.1)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_1|I)=\frac{0.015}{0.015+0.12+0.03}$

$P(s_1|I)=\frac{0.015}{0.165}$

$P(s_1|I)=\frac{1}{11}$

Therefore the value of P(s₁|I) is $\frac{1}{11}$ .

(2)

We need to find the value of P(s₂|I).

$P(s_2|I)=\frac{P(I|s_2)P(s_2)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_2|I)=\frac{(0.2)(0.6)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_2|I)=\frac{0.12}{0.015+0.12+0.03}$

$P(s_2|I)=\frac{0.12}{0.165}$

$P(s_2|I)=\frac{8}{11}$

Therefore the value of P(s₂|I) is $\frac{8}{11}$ .

(3)

We need to find the value of P(s₃|I).

$P(s_3|I)=\frac{P(I|s_3)P(s_3)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_3|I)=\frac{(0.1)(0.3)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_3|I)=\frac{0.03}{0.015+0.12+0.03}$

$P(s_3|I)=\frac{0.03}{0.165}$

$P(s_3|I)=\frac{2}{11}$

Therefore the value of P(s₃|I) is $\frac{2}{11}$ .

4 0

2 years ago

How to find the greatest common factor of 24 and 90?

love history [14]

Well, they're both divisible by 2 (24 ÷ 2 = 12) (90 ÷ 2 = 45)

3 0

2 years ago

The function h(x) is defined as shown.

belka [17]

Hello,

h(x)= if x<3 then x+2
else -x+8

(–∞5[ U [5 –∞)=(–∞ 5]

Answer B

7 0

2 years ago

Read 2 more answers