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2 years ago

Show work please!!!!! HELPPPP

Mathematics

1 answer:

scoundrel [369]2 years ago

7 0

If you would like to find the volume here:

L•W•H

Big Rectangular: 10•14•6
Small Rectangular: 1•2•3

Surface Area:

2(L•W) + 2(L•H) + 2(W•H)

Big Rectangular: 2(10•14) + 2(10•6) + 2(14•6)
Small Rectangular: 2(1•2) + 2(1•3) + 2(2•3)

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Amanda reads 15 pages of her 150 page book in 2 hours at this rate how long will it take her to read the entire book

Crazy boy [7]

Ratio and proportion 15 pages/2hours=150 pages/x hours cross multiply 15x= 300 x-20 it will take Amanda 20 hours to read the entire book

3 0

3 years ago

Pamela Is 10 years older than Jiri. The sum of their ages is 96. What is Jiri’s age

Solnce55 [7]

43 is jiri and 53 is pamela

8 0

3 years ago

Read 2 more answers

Molly Invests $8,600 into her son's college fund, which earns 3% annual interest

8_murik_8 [283]

Answer:

n= 13.8 years

Step-by-step explanation:

First, we need to determine the daily interest rate:

Daily interest rate= 0.03 / 365

Daily interest rate= 0.000082

Now, using the following formula we can determine the number of days and years:

n= ln(FV/PV) / ln(1+i)

n= ln(13,000 / 8,600) / ln(1.000082)

n= 5,039 days

In years:

n= 5,039/365

n= 13.8 years

7 0

2 years ago

Write a rule for the nth term of the arithmetic sequence: a11= 50, d = 7

ivolga24 [154]

Answer:

Required rule for $n^{th}$ is $a_{n}=7n-27$ .

Step-by-step explanation:

Given that,

$a_{11} = 50,\ \ d=7$

From the question: we have to write the $n^{th}$ term of Arithmetic sequence.

Arithmetic Sequence or Arithmetic progression (A.P) : It is a sequence which possess that difference between of two successive sequence is always constant.

$a_{1} ,a_{2},a_{3},a_{4}.....................a_{n-1},a_{n}$

where, $a_{1}$ is the first term of A.P

$d$ is the common difference.

$a_{n}$ is the last term or general term.

The above sequence to be in A.P then their common difference should be equal.

$d=a_{2}-a_{1} =a_{3}-a_{2}=a_{4} -a_{3} ..........................a_{n}-a_{n-1}$

Now, Formula of General Term is $a_{n}=a+(n-1)d$

So, $a_{11}= a+(11-1)d\\ a_{11} = a+10d$

Substituting the value of $a_{11} = 50,\ \ d=7$ we get,

$50=a+10\times7\\50=a+70\\a=-20$

Then General term ( $a_{n}$ ) of given data is

$a_{n}=-20+(n-1)7\\a_{n}=-20+7n-7\\a_{n}=7n-27$

Therefore, Required rule for $n^{th}$ is $a_{n}=7n-27$ .

5 0

3 years ago

Based on a sample of 100 employees a 95% confidence interval is calculated for the mean age of all employees at a large firm. Th

babymother [125]

Answer:

$\= x = 40.85$

$E = 5.85$

$t_c = 2.08$

$t_c = 1.282$

Explanation:

From the question we are told that

The sample size is n = 100

The upper limit of the 95% confidence interval is b = 47.2 years

The lower limit of the 95% confidence interval is a = 34.5 years

Generally the sample mean is mathematically represented as

$\= x = \frac{a + b }{2}$

=> $\= x = \frac{47.2 + 34.5 }{2}$

=> $\= x = 40.85$

Generally the margin of error is mathematically represented as

$E = \frac{b- a }{ 2}$

=> $E = \frac{47.2- 34.5 }{ 2}$

=> $E = 5.85$

Considering question C a

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

The sample size is n = 22

Given that the sample size is not sufficient enough i.e $n < 30$ we will make use of the student t distribution table

Generally the degree of freedom is mathematically represented as

$df = n- 1$

=> $df = 22 - 1$

=> $df = 21$

Generally from the student t distribution table the critical value of $\frac{\alpha }{2}$ at a degree of freedom of 21 is

$t_c =t_{\frac{\alpha }{2} , 21 } = 2.08$

Considering question C b

From the question we are told the confidence level is 80% , hence the level of significance is

$\alpha = (100 - 80 ) \%$

=> $\alpha = 0.20$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$t_c =Z_{\frac{\alpha }{2} } = 1.282$

4 0

2 years ago