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notka56 [123]
2 years ago
7

Brice has $1200 in the bank. He wants to save a total of $3000 by depositing $40 per week from his paycheck. Write and use an eq

uation to find how many weeks he needs to reach his goal.

Mathematics
1 answer:
Anon25 [30]2 years ago
6 0

$$3000-$1200=$1800

take $1800÷$40=45

<h3>answ=45 weeks</h3>

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A research analyst for a car rental agency found that the probability distribution function for the age of the rental car (in ye
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Answer:

10 years

Step-by-step explanation:

Given

Decline from (0,h) to (10,0)

See attachment for graph

Required

Determine the age when height = 0

The function of the graph is represented as: (age, height)

So, we need to read from the graph the corresponding value of age (on the x-axis) when height = 0 (i.e. the y-axis)

From the attached graph;

age = 10 when height = 0

<em>Hence, the age of the car is 10 years</em>

4 0
2 years ago
We created this inequality to represent how Amy can meet her revenue goals:
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Answer:

x = 2

Price = $29

Step-by-step explanation:

If x is the number of $2 dollars increases, then the minimum number of increases required to meet her goal is:

-100x^2 + 15,750x + 212,500 \geq 243,600\\-100x^2 + 15,750x -31,100\geq 0\\x^2 -157.5x + 311\geq 0\\x=\frac{157.5\pm\sqrt{157.5^2-(4*1*311)} }{2}\\ x_1= 2\\x_2=155.5

The only realistic value is x = 2 $2 increases

If the original price was $25, then the new rice required to reach Amy's goal is:

P=25+2*2\\P=\$29

The minimum ticket price is $29.

7 0
3 years ago
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Add the equations 2x-3y=-1 + 3x+3y=26 A. 5x-6y=-27 B. 5x=27 C. 6y=25 D. 5x=25​
jenyasd209 [6]

Answer:

D) 5x = 25

Step-by-step explanation:

-1+26 = 25

2x + 3x = 5x

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6 0
3 years ago
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Which of the following appear in the diagram below?
mote1985 [20]

Answer:

Option B ; C ; D

Option B i.e YX

Option C i.e YW

Option D i.e <XYZ

6 0
3 years ago
In a certain population of the eastern thwump bird, the wingspan of the individual birds follows an approximately normal distrib
Greeley [361]

Answer:

a) P(48 < x < 58) = 0.576

b) P(X ≥ 1) = 0.9863

c) E(X) 2.88

d) P(x < 48) = 0.212

e) P(X > 2) = 0.06755

Step-by-step explanation:

The mean of the wingspan of the birds = μ = 53.0 mm

The standard deviation = σ = 6.25 mm

a) Probability of a bird having a wingspan between 48 mm and 58 mm can be found by modelling the problem as a normal distribution problem.

To solve this, we first normalize/standardize the two wingspans concerned.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ

For wingspan 48 mm

z = (48 - 53)/6.25 = - 0.80

For wingspan 58 mm

z = (58 - 53)/6.25 = 0.80

To determine the probability that the wingspan of the first bird chosen is between 48 and 58 mm long. P(48 < x < 58) = P(-0.80 < z < 0.80)

We'll use data from the normal probability table for these probabilities

P(48 < x < 58) = P(-0.80 < z < 0.80) = P(z < 0.8) - P(z < -0.8) = 0.788 - 0.212 = 0.576

b) The probability that at least one of the five birds has a wingspan between 48 and 58 mm = 1 - (Probability that none of the five birds has a wingspan between 48 and 58 mm)

P(X ≥ 1) = 1 - P(X=0)

Probability that none of the five birds have a wingspan between 48 and 58 mm is a binomial distribution problem.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of birds = 5

x = Number of successes required = number of birds with wingspan between 48 mm and 58 mm = 0

p = probability of success = Probability of one bird having wingspan between 48 mm and 58 mm = 0.576

q = probability of failure = Probability of one bird not having wingspan between 48 mm and 58 mm = 1 - 0.576 = 0.424

P(X=0) = ⁵C₀ (0.576)⁰ (1 - 0.576)⁵ = (1) (1) (0.424)⁵ = 0.0137

The probability that at least one of the five birds has a wingspan between 48 and 58 mm = P(X≥1) = 1 - P(X=0) = 1 - 0.0137 = 0.9863

c) The expected number of birds in this sample whose wingspan is between 48 and 58 mm.

Expected value is a sum of each variable and its probability,

E(X) = mean = np = 5×0.576 = 2.88

d) The probability that the wingspan of a randomly chosen bird is less than 48 mm long

Using the normal distribution tables again

P(x < 48) = P(z < -0.8) = 1 - P(z ≥ -0.8) = 1 - P(z ≤ 0.8) = 1 - 0.788 = 0.212

e) The probability that more than two of the five birds have wingspans less than 48 mm long = P(X > 2) = P(X=3) + P(X=4) + P(X=5)

This is also a binomial distribution problem,

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of birds = 5

x = Number of successes required = number of birds with wingspan less than 48 mm = more than 2 i.e. 3,4 and 5.

p = probability of success = Probability of one bird having wingspan less than 48 mm = 0.212

q = probability of failure = Probability of one bird not having wingspan less than 48 mm = 1 - p = 0.788

P(X > 2) = P(X=3) + P(X=4) + P(X=5)

P(X > 2) = 0.05916433913 + 0.00795865476 + 0.00042823218

P(X > 2) = 0.06755122607 = 0.06755

5 0
3 years ago
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