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Allisa [31]
2 years ago
9

A certain forest covers an area of 2100 km^2. Suppose that each year this area decreases by 5.25%. What will the area be after 6

years
Mathematics
1 answer:
Neko [114]2 years ago
4 0

Answer:

The area after 6 years is 1519.49km^2

<em></em>

Step-by-step explanation:

Given

a = 2100 --- Initial Value

r = 5.25\% --- Rate

Required

Determine the area after 6 years

Because the area reduces,, the following formula will be used to solve the question.

y = a*(1 - r)^t

Here

t = 6

So, we have:

y = 2100*(1 - 5.25\%)^6

Convert percentage to decimal

y = 2100*(1 - 0.0525)^6

y = 2100*(0.9475)^6

y = 1519.47866516

y = 1519.49 -- approximated

<em>The area after 6 years is </em>1519.49km^2<em></em>

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3 years ago
A museum opened at 9:00 a.m. In the first hour, 350 people purchased admission tickets. In the second hour, 20% more people purc
Furkat [3]

Answer:

13,475

Step-by-step explanation:

First hour

350 people at $17.50 per ticket

350* $17.50 =$6125

Second hour

We add 20% more

350+350*.20 =350+70 =420

There were 420 people at $17.50

420*$17.50 =$7350

Add the total for the 2 hours

6125+7350=13,475

3 0
3 years ago
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If, in 7 years, Susan will be 2 times as old as she was 3 years ago, what is Susan’s present age? PLEAZE EXPLAIN HOW AND DO IT W
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Let Susan's present age be x

Now, after 7 years ( her age will become 7 +x then) she'll be 2 as old she was 3 years ago.

Her age 3 years ago must be x-3

Now according to question,

twice Her age 3 years ago = her age after 7 years

now,

2(x-3) = 7+x
2x-6 = 7+x
2x-x =7+6
x = 13
x = 13



Therefore, her age presenr age is 13 years.
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3 years ago
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Help me please. need quickly. What is the value of Z
larisa86 [58]

Answer:

z = 6

Step-by-step explanation:

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8 0
2 years ago
In a survey, 11 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped
pickupchik [31]

Answer:

The population standard deviation is not known.

90% Confidence interval by T₁₀-distribution: (38.3, 53.7).

Step-by-step explanation:

The "standard deviation" of $14 comes from a survey. In other words, the true population standard deviation is not known, and the $14 here is an estimate. Thus, find the confidence interval with the Student t-distribution. The sample size is 11. The degree of freedom is thus 11 - 1 = 10.

Start by finding 1/2 the width of this confidence interval. The confidence level of this interval is 90%. In other words, the area under the bell curve within this interval is 0.90. However, this curve is symmetric. As a result,

  • The area to the left of the lower end of the interval shall be 1/2 \cdot (1 - 0.90)= 0.05.
  • The area to the left of the upper end of the interval shall be 0.05 + 0.90 = 0.95.

Look up the t-score of the upper end on an inverse t-table. Focus on the entry with

  • a degree of freedom of 10, and
  • a cumulative probability of 0.95.

t \approx 1.812.

This value can also be found with technology.

The formula for 1/2 the width of a confidence interval where standard deviation is unknown (only an estimate) is:

\displaystyle t \cdot \frac{s_{n-1}}{\sqrt{n}},

where

  • t is the t-score at the upper end of the interval,
  • s_{n-1} is the unbiased estimate for the standard deviation, and
  • n is the sample size.

For this confidence interval:

  • t \approx 1.812,
  • s_{n-1} = 14, and
  • n = 11.

Hence the width of the 90% confidence interval is

\displaystyle 1.812 \times \frac{14}{\sqrt{10}} \approx 7.65.

The confidence interval is centered at the unbiased estimate of the population mean. The 90% confidence interval will be approximately:

(38.3, 53.7).

5 0
3 years ago
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