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3 years ago

PLEASE HELP I AM DESPERATE I WILL GIVE THANKS, 5 STARS, AND THE BRAINLIEST!!!

Mathematics

1 answer:

Amanda [17]3 years ago

3 0

g(-2) + g(2) = 22

Step-by-step explanation:

For x = -2, g(x) = x^2 - 3x

g(-2) = (-2)^2 - 3(-2) = 4 + 6 = 10

For x = 2, g(x) = 12, therefore

g(-2) + g(2) = 10 + 12 = 22

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How much would you need to deposit in an account now in order to have $5000 in the account in 10 years? Assume the account earns

Blababa [14]

Answer:

$12500

Step-by-step explanation:

use si formula

SI=ptr/100

si=5000

t=10

r=4

6 0

3 years ago

Please help me! ASAP!!!!!

Greeley [361]

Answer:

x=36

Step-by-step explanation:

17+3 =20

now we have :

x+4

___ = 20

20*2= 40

x+4=40

x=36

hopes this helps

4 0

3 years ago

No link or bot answer the question

sasho [114]

Answer:

c. Rational

because → whole numbers are positive and Starts from 0 so not whole number

Integers can be negative and positive

Examples of numbers that are not integers are: -1.43, 1 3/4, 3.14, . 09, and 5,643.1.

Irrational number that cannot be written in p/q

but -2.4 written in p/q = -24/100

<h3>so c. Rational Number is correct</h3>

3 0

2 years ago

Solve arctan(-sqrt(3))

tatyana61 [14]

We are given with the expression arctan(-sqrt(3)) and asked to evaluate it. In this case, we can use a calculator or the rule of common triangles to answer this question. the value of <span>arctan(-sqrt(3)) is -60. Since negative tan is found in 2nd and 4th quadrant, the angles are 180-60 or 120 degrees and 360-60 or 300 degrees.</span>

6 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