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3 years ago

7

Diana invested $3000 in a savings account for 3 years. She earned $450 in interest over that time period. What interest rate did

she earn? Use the formula I=Prt to find your answer, where I is interest, P is principal, r is rate and t is time. Enter your solution in decimal form rounded to the nearest hundredth. For example, if your solution is 12%, you would enter 0.12.

2 answers:

alukav5142 [94]3 years ago

7 0

Answer:

R=0.05

Step-by-step explanation:

Mrac [35]3 years ago

5 0

R=i/pt
R=450÷(3,000×3)
R=0.05

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valkas [14]

Answer:

x = 2

Step-by-step explanation:

First, let's write out our equation:

$2x-4=6-3x$

I want to isolate x on one side, so first, I'll add 4 to both sides to remove the -4 from next to 2x:

$2x-4+4=6-3x+4\\2x=-3x + 10$

Notice that I combined like terms with the -4 and 4 (to get 0) and the 6 and 4 on the right side (to get 10). Next, I'll add 3x to both sides:

$2x + 3x = -3x + 3x + 10$

And then I'll add like terms:

$5x=10$

Now, all we have to do is divide both sides by 5:

$5x/5=10/5\\x=2$

And there's our answer. Hopefully that's helpful! :)

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2 years ago

Can someone help me with this! Ty

rosijanka [135]

Answer:

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Step-by-step explanation:

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3 years ago

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Twenty percent of drivers driving between 10 pm and 3 am are drunken drivers. In a random sample of 12 drivers driving between 1

Lesechka [4]

Answer:

(a) 0.28347

(b) 0.36909

(c) 0.0039

(d) 0.9806

Step-by-step explanation:

Given information:

n=12

p = 20% = 0.2

q = 1-p = 1-0.2 = 0.8

Binomial formula:

$P(x=r)=^nC_rp^rq^{n-r}$

(a) Exactly two will be drunken drivers.

$P(x=2)=^{12}C_{2}(0.2)^{2}(0.8)^{12-2}$

$P(x=2)=66(0.2)^{2}(0.8)^{10}$

$P(x=2)=\approx 0.28347$

Therefore, the probability that exactly two will be drunken drivers is 0.28347.

(b)Three or four will be drunken drivers.

$P(x=3\text{ or }x=4)=P(x=3)\cup P(x=4)$

$P(x=3\text{ or }x=4)=P(x=3)+P(x=4)$

Using binomial we get

$P(x=3\text{ or }x=4)=^{12}C_{3}(0.2)^{3}(0.8)^{12-3}+^{12}C_{4}(0.2)^{4}(0.8)^{12-4}$

$P(x=3\text{ or }x=4)=0.236223+0.132876$

$P(x=3\text{ or }x=4)\approx 0.369099$

Therefore, the probability that three or four will be drunken drivers is 0.3691.

(c)

At least 7 will be drunken drivers.

$P(x\geq 7)=1-P(x$

$P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)]$

$P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155]$

$P(x\leq 7)=1-[0.9961]$

$P(x\leq 7)=0.0039$

Therefore, the probability of at least 7 will be drunken drivers is 0.0039.

(d) At most 5 will be drunken drivers.

$P(x\leq 5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)$

$P(x\leq 5)=0.06872+0.20616+0.28347+0.23622+0.13288+0.05315$

$P(x\leq 5)=0.9806$

Therefore, the probability of at most 5 will be drunken drivers is 0.9806.

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3 years ago

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Serga [27]

Answer:

I think that the answer isssss B

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3 years ago

Select the correct answer.

ivolga24 [154]

I think the answer is c

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1 year ago

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