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1 year ago

Evaluate the expression 2.1 + (–3.7h) + 1.9h – 1.4

Mathematics

1 answer:

zubka84 [21]1 year ago

5 0

2.1 + (-3.7h) + 1.9h - 1.4
2.1 - 3.7h + 1.9h - 1.4
2.1 - 1.8h - 1.4
0.7 - 1.8h

the simplified form is 0.7 - 1.8h

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Which model represents the factors of 4x2 – 9? An algebra tile configuration. 5 tiles are in the Factor 1 spot: 2 +x , 3 negativ

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Answer:

Step-by-step explanation:

4x² - 9

(2x)² - 3²

(2x + 3)(2x - 3)

Factor 1 spot:

2 tiles of x

3 tiles of positive 1

Factor 2 spot:

2 tiles of x

3 tiles of negative 1

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Grace is comparing cell phone plans. A prepaid phone plan costs $0.20 per minute and has no monthly fee. A contracted phone plan

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Carolyn is using the table to find 360% of 15. What values do X and Y represent in her table? Percent Total 100% 100% 100% 20% 2

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Answer: c

Step-by-step explanation:

Based on the percent breakdown of 360%.

100% 100% 100% 20%20%20%

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2 years ago

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A bank gets an average of 12 customers per hour. Assume the variable follows a Poisson distribution. Find the probability that t

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Answer:

75.76% probability that there will be 10 or more customers at this bank in one hour.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

A bank gets an average of 12 customers per hour.

This means that $\mu = 12$

Find the probability that there will be 10 or more customers at this bank in one hour.

Either there are less than 10 customers, or there are 10 or more. The sum of the probabilities of these events is 1. Then

$P(X < 10) + P(X \geq 10) = 1$

We want $P(X \geq 10)$

Then

$P(X \geq 10) = 1 - P(X < 10)$

In which

$P(X < 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-12}*12^{0}}{(0)!} \approx 0$

$P(X = 1) = \frac{e^{-12}*12^{1}}{(1)!} = 0.0001$

$P(X = 2) = \frac{e^{-12}*12^{2}}{(2)!} = 0.0004$

$P(X = 3) = \frac{e^{-12}*12^{3}}{(3)!} = 0.0018$

$P(X = 4) = \frac{e^{-12}*12^{4}}{(4)!} = 0.0053$

$P(X = 5) = \frac{e^{-12}*12^{5}}{(5)!} = 0.0127$

$P(X = 6) = \frac{e^{-12}*12^{6}}{(6)!} = 0.0255$

$P(X = 7) = \frac{e^{-12}*12^{7}}{(7)!} = 0.0437$

$P(X = 8) = \frac{e^{-12}*12^{8}}{(8)!} = 0.0655$

$P(X = 9) = \frac{e^{-12}*12^{9}}{(9)!} = 0.0874$

$P(X < 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) = 0 + 0.0001 + 0.0004 + 0.0018 + 0.0053 + 0.0127 + 0.0255 + 0.0437 + 0.0655 + 0.0874 = 0.2424$

Then

$P(X \geq 10) = 1 - P(X < 10) = 1 - 0.2424 = 0.7576$

75.76% probability that there will be 10 or more customers at this bank in one hour.

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2 years ago

Rita promises her daughter on her 12th birthday that she will give her $12,000 for college on her 18th birthday. How much does R

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Answer:

She should invest 6976.74 $ principal on daughter's 12th birthday.

Explanation:

We have I = prt, where I represents simple interest, p represents principal, r represents interest rate, and t represents time in years.

Here p + I = 12,000 $

So, I = 12000-p

Interest rate r = 12% =0.12

Number of years = 18 - 12 = 6 years.

Substituting

12000-p = p x 0.12 x 6

p + 0.72 p = 12000

1.72 p = 12000

p = 6976.74 $

So, she should invest 6976.74 $ principal on daughter's 12th birthday.

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