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3 years ago

9

Last friday trevon had $29. over the weekend he received some money for cleaning the attic. he now has $41. how much money did h

e receive?

2 answers:

snow_lady [41]3 years ago

6 0

We can subtract $41 from $29

41-29=12

Trevon will receive $12

Agata [3.3K]3 years ago

5 0

$41 - $29 = your answer

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Write a proof for the biconditional using the inverse and contrapositive.

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I don’t really understand the question at hand but let me try to help

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3 years ago

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2 years ago

Carla wants to start a college fund for her daughter Lila. She puts $63,000 into an account that grows at a rate of 2.55% per ye

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Answer:

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1+r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = 63000

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n = 12 because it was compounded 12 times in a year.

Therefore, function, C(t), that represents the amount of money in the account t years after the account is opened is

C(t) = 63000(1 + 0.0255/12)^12t

C(t) = 63000(1.002125)^12t

For C(t) = 100000,

100000 = 63000(1.002125)^12t

100000/63000 = (1.002125)^12t

1.587 = 1.002125)^12t

Taking log of both sides

Log 1.587 = log 1.002125)^12t

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4 years ago

The sum of 5.5 and twice a number is -10

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The answer is n= 7.75.

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2 years ago

Read 2 more answers

NarStor, a computer disk drive manufacturer, claims that the median time until failure for their hard drives is more than 14,400

Ne4ueva [31]

Answer:

The test statistics is $t = -1.727$

Step-by-step explanation:

From the question we are told that

The data given is

330 620 1870 2410 4620 6396 7822 81028309 12882 14419 16092 18384 20916 23812 25814

The population mean is $\mu = 14400$

The sample size is n = 16

The null hypothesis is $\mu \le 14400$

The alternative hypothesis is $H_a : \mu > 14400$

The sample mean is mathematically evaluated as

$\= x = \frac{\sum x_i}{n}$

So

$\= x = \frac{330+ 620+ 1870 +2410+ 4620+ 6396+ 7822+ 8102+8309+ 12882+ 14419+ 16092+ 18384 +20916+ 23812+ 25814 }{16}$

=> $\= x = 10799.9$

The standard deviation is mathematically represented as

$\sigma =\sqrt{\frac{ \sum (x_i - \=x)^2}{n}}$

So

$\sigma =\sqrt{\frac{(330- 10799.9)^2 + (620- 10799.9)^2+ (1870- 10799.9)^2 +(2410- 10799.9)^2 + (4620- 10799.9)^2 +(6396- 10799.9)^2 +(7822- 10799.9)^2 }{16}} \ ..$

$..\sqrt{ \frac{(8102 - 10799.9)^2 +(8309 - 10799.9)^2 + (12882 - 10799.9)^2 + (14419 - 10799.9)^2 + (16092 - 10799.9)^2 + (18384 - 10799.9)^2 +(20916 - 10799.9)^2 }{16}} \ ...$

$\ ... \sqrt{\frac{(23812 - 10799.9)^2 +(25814 - 10799.9)^2 }{16}}$

=> $\sigma = 8340$

Generally the test statistic is mathematically represented as

$t = \frac{10799.9- 14400}{ \frac{8340}{\sqrt{16} } }$

$t = -1.727$

From the z-table the p-value is

$p-value = P(Z > t) = P(Z > -1.727) = 0.95792$

From the values obtained we see that

$p-value > \alpha$ so we fail to reject the null hypothesis

Which implies that the claim of the NarStor is wrong

5 0

3 years ago