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alexandr402 [8]

2 years ago

8

A field is a rectangle with a perimeter of 1260 feet. The length is 100 feet more than width. Find the width and length of the r

ectangle field.

1 answer:

lyudmila [28]2 years ago

5 0

To solve this problem you must apply the proccedure shown below:

1. The formula for calculate the perimeter of a rectangle is:

$P=2L+2W$

Where $L$ is the length and $W$ is the width.

2. The length is $100ft$ more than the width, therefore:

$L=W+100$

3. Substitute values into the formula for calculate the perimeter and solve for the width:

$1260=2(W+100)+2W\\ W=265ft$

3. Then, the length is:

$L=265ft+100ft\\ L=365ft$

The answer is: The length is $365ft$ and the width is $265ft$

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1. The rule for an input/output table is a = w+ 7. What is the output if the input is 3?

natta225 [31]

I am not sure but random guess is 0.056

7 0

2 years ago

Rewa Delta Union Rugby CEO has become concerned about the slow pace of the rugby games played in the current union rugby, fearin

ozzi

Answer:

We conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.

Step-by-step explanation:

We are given that Before the meeting, the mean duration of the 15-sided rugby game time was 3 hours, 5 minutes, that is, 185 minutes.

A random sample of 36 of the 15-sided rugby games after the meeting showed a mean of 179 minutes with a standard deviation of 12 minutes.

Let $\mu$ = <em><u>mean duration of 15-sided union rugby games after the meeting.</u></em>

So, Null Hypothesis, $H_0$ : $\mu \geq$ 185 minutes {means that the mean duration of 15-sided union rugby games has increased or remained same after the meeting}

Alternate Hypothesis, $H_A$ : $\mu$ < 185 minutes {means that the mean duration of 15-sided union rugby games has decreased after the meeting}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean duration of 15-sided union rugby games = 179 min

s = sample standard deviation = 12 minutes

n = sample of 15-sided rugby games = 36

So, <u><em>the test statistics</em></u> = $\frac{179-185}{\frac{12}{\sqrt{36} } }$ ~ $t_3_5$

= -3

The value of t test statistics is -3.

<u>Now, at 1% significance level the t table gives critical value of -2.437 at 35 degree of freedom for left-tailed test.</u>

Since our test statistic is less than the critical value of t as -3 < -2.437, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.

6 0

3 years ago

The product of 3 and 1 less than a number is

MAVERICK [17]

Answer: 4 is lesser than the all of the numbers above it.

Step-by-step explanation: I don’t really understand what you are trying to ask, Maybe re-word what you were trying to say?

8 0

3 years ago

The old time train museum is raising the price of tickets for train rides.Tickets used to cost $4.80 each. The new ticket price

SVETLANKA909090 [29]

Only add $4.80 + $1.25

5 0

3 years ago

Read 2 more answers

Kristen invests $ 5745 in a bank. The bank 6.5% interest compounded monthly. How long must she leave the money in the bank for i

podryga [215]

Answer:

1. 10.7 years

2. 17.0 years

3. 2nd option

Step-by-step explanation:

Use formula for compounded interest

$A=P\cdot \left(1+\dfrac{r}{n}\right)^{nt},$

where

A is final value, P is initial value, r is interest rate (as decimal), n is number of periods and t is number of years.

In your case,

1. P=$5745, A=2P=$11490, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then

$11490=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},\\ \\2=(1.0054)^{12t},\\ \\12t=\log_{1.0054}2,\\ \\t=\dfrac{1}{12}\log_ {1.0054}2\approx 10.7\ years.$

2. P=$5745, A=3P=$17235, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then

$17235=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},$

$3=(1.0054)^{12t},$

$12t=\log_{1.0054}3,$

$t=\dfrac{1}{12}\log_{1.0054}3\approx 17.0\ years.$

3. <u>1 choice:</u> P=$5745, n=12 (compounded monthly), r=0.065 (6.5%), t=5 years and A is unknown. Then

$A=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12\cdot 5},\\ \\A=5745\cdot (1.0054)^{60}\approx \$7936.39.$

<u>2 choice:</u> P=$5745, n=4 (compounded monthly), r=0.0675 (6.75%), t=5 years and A is unknown. Then

$A=5745\cdot \left(1+\dfrac{0.0675}{4}\right)^{4\cdot 5},\\ \\A=5745\cdot (1.0169)^{20}\approx \$8032.58.$

The best will be 2nd option, because $8032.58>$7936.39

5 0

2 years ago

Read 2 more answers