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3 years ago

PRONTO PLS-------

Mathematics

1 answer:

hram777 [196]3 years ago

8 0

Answer:

t = 44 years

Step-by-step explanation:

t = (1/r)(I/P-1)

t = (1/0.05) {(8000/2500) - 1}

t = 44

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Y=2x+7 3x-y=-9 Solve each system of equation by the substitution method.

givi [52]

Plug in the first equation into the he second equation

3x-(2x+7)=-9
3x-2x-7=-9
x-7=-9
x=-2

plug in x=-2 into the first equation

y=2(-2)+7
y=-4+7
y=3

answer: (-2,3)

4 0

3 years ago

Read 2 more answers

Use the function table to determine the slope.EXPLAIN YOUR THINKING AND HOW YOU GOT YOUR ANSWER.I WILL MARK YOU BRAINLIST!

Tom [10]

Answer:

3¼

Step-by-step explanation:

$formula \frac{{y}^{2} - {y}^{1} }{ {x}^{2} - {x}^{1} }$

We take any of the two points shown in the question. I will take (-3, 1) and (1, 14).

x¹ = -3

y¹ = 1

x² = 1

y² = 14

Now, we sub these figures into the formula.

$\frac{14 - 1}{1 - ( - 3)}$

This leaves us with 14-1/1+3, which we can make into 13/4

13/4 = 3¼

Disclaimer: It is not actually x (or y) in the power of two, but a way to distinguish one x (or y) from the other. It is a label of sorts.

3 0

3 years ago

Read 2 more answers

|-9c|=8 and solve for c

melisa1 [442]

C = negative 8/9 or 8/9

5 0

3 years ago

the perimeter of a rectangular garden is 52 feet the length of the garden is 4 feet less than twice the width what is the width

777dan777 [17]

The width would be 10ft and the length would be 16ft

when you multiply 10 by 2 and 16 by 2 you get 20 and 32 and when those are added it gets 52

Hope that helps

If you have any questions leave a comment

7 0

3 years ago

he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and s

sp2606 [1]

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

$P(X$

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) = $[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]$

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

$NORMSDIST(6.366)-NORMSDIST(-1.47)$

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) = $P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]$

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

= 0.0000

3 0

3 years ago