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grigory [225]
3 years ago
14

Simplify the expression.

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
7 0

Answer: 8nx−7p−6x

Step-by-step explanation:

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115. Sami opens an account and deposits $100 into it at the end of each month. The account earns 2% per year compounded monthly.
givi [52]

Step-by-step explanation:

Geometric series.

Month 3.

      100(1+\frac{0.02}{12})^2 + 100(1+\frac{0.02}{12})+100

Month 4.

100( 1 + \frac{0.02}{12})^3 + 100( 1 + \frac{0.02}{12})^2+100(1+\frac{0.02}{12})+100

Month 5.

100(1 + \frac{0.02}{12})^4 + 100(1 + \frac{0.02}{12})^3 + 100(1 + \frac{0.02}{12})^2 +100( 1 + \frac{0.02}{12}) + 100

4 0
4 years ago
"An ordinance requiring that a smoke detector be installed in all previously constructed houses has been in effect in a particul
Galina-37 [17]

Answer:

a) Probability that the claim is rejected when the actual value of p is 0.8 = P(X ≤ 15) = 0.0173

b) Probability of not rejecting the claim when p = 0.7, P(X > 15) = 0.8106

when p = 0.6, P(X > 15) = 0.4246

c) Check Explanation

The error probabilities are evidently lower when 15 is replaced with 14 in the calculations.

Step-by-step explanation:

p is the true proportion of houses with smoke detectors and p = 0.80

The claim that 80% of houses have smoke detectors is rejected if in a sample of 25 houses, not more than 15 houses have smoke detectors.

If X is the number of homes with detectors among the 25 sampled

a) Probability that the claim is rejected when the actual value of p is 0.8 = P(X ≤ 15)

This is a binomial distribution problem

A binomial experiment is one in which the probability of success doesn't change with every run or number of trials (probability that each house has a detector is 0.80)

It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure (we are sampling 25 houses with each of them either having or not having a detector)

The outcome of each trial/run of a binomial experiment is independent of one another.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 25 houses sampled

x = Number of successes required = less than or equal to 15

p = probability of success = probability that a house has smoke detectors = 0.80

q = probability of failure = probability that a house does NOT have smoke detectors = 1 - p = 1 - 0.80 = 0.20

P(X ≤ 15) = Sum of probabilities from P(X = 0) to P(X = 15) = 0.01733186954 = 0.01733

b) Probability of not rejecting the claim when p= 0.7 when p= 0.6

For us not to reject the claim, we need more than 15 houses with detectors, hence, th is probability = P(X > 15), but p = 0.7 and 0.6 respectively for this question.

n = total number of sample spaces = 25 houses sampled

x = Number of successes required = more than 15

p = probability that a house has smoke detectors = 0.70, then 0.60

q = probability of failure = probability that a house does NOT have smoke detectors = 1 - p = 1 - 0.70 = 0.30

And 1 - 0.60 = 0.40

P(X > 15) = sum of probabilities from P(X = 15) to P(X = 25)

When p = 0.70, P(X > 15) = 0.8105639765 = 0.8106

When p = 0.60, P(X > 15) = 0.42461701767 = 0.4246

c) How do the "error probabilities" of parts (a) and (b) change if the value 15 in the decision rule is replaced by 14.

The error probabilities include the probability of the claim being false.

When X = 15

(Error probability when p = 0.80) = 0.0173

when p = 0.70, error probability = P(X ≤ 15) = 1 - P(X > 15) = 1 - 0.8106 = 0.1894

when p = 0.60, error probability = 1 - 0.4246 = 0.5754

When X = 14

(Error probability when p = 0.80) = P(X ≤ 14) = 0.00555

when p = 0.70, error probability = P(X ≤ 14) = 0.0978

when p = 0.60, error probability = P(X ≤ 14) = 0.4142

The error probabilities are evidently lower when 15 is replaced with 14 in the calculations.

Hope this Helps!!!

6 0
3 years ago
Ramon is a waiter. He waits on a table of 4 whose bill comes to $99.69. If Ramon receives a 20% tip, approximately how much will
vlabodo [156]
99.69/100=.9969
.9969*20=19.938 
Ramon would receive ~ $19.94
4 0
3 years ago
Given: // CDAB , ∠≅∠ DB and ≅ EDBF . Prove: Δ ABF ≅ Δ CED
SCORPION-xisa [38]

Given: line segment AB // to line segment CD, ∠B ≅∠D and line segment BF ≅ to line segment ED. Prove: Δ ABF ≅ Δ CED.

Follow the matching numbers on the statement versus reason chart.

Statement:

1. line segment AB // to line segment CD.

2. ∠B ≅∠D

3. line segment BF ≅ to line segment ED.

4. ∠A ≅∠C

5. Δ ABF ≅ Δ CED

 

Reason:

1. Given

2. Given

3. Given

4. Alternate interior angles are congruent.

5. Corresponding parts of congruent triangles are congruent.

4 0
4 years ago
What percentage of the population voted for trump?
Vilka [71]
22.5% voted for trump for president
7 0
3 years ago
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