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kow [346]
2 years ago
10

C) Your parents have been advised to save 5% of their income for your college education

Mathematics
2 answers:
My name is Ann [436]2 years ago
7 0

48,948 * 5% = 2447.4

MatroZZZ [7]2 years ago
4 0

Answer:

Your parents will have saved $2447.40 in a year.

Step-by-step explanation:

Your parents have been advised to save 5% of their income for your college education.

Their annual income is = $48948

So, 5% of this will be = 0.05\times48948=2447.40 dollars

Hence, your parents will have saved $2447.40 in a year.

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PIT_PIT [208]
We have that
<span>K = C + 273.15
solve for C
subtract 273.15 both sides
K-273.15=C
C=K-273.15

the answer is
</span>C=K-273.15<span>


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5 0
2 years ago
Read 2 more answers
A charity receives 2025 contributions. Contributions are assumed to be mutually independent and identically distributed with mea
uysha [10]

Answer:

The 90th percentile for the distribution of the total contributions is $6,342,525.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For sums of size n, the mean is \mu*n and the standard deviation is s = \sqrt{n}*\sigma

In this question:

n = 2025, \mu = 3125*2025 = 6328125, \sigma = \sqrt{2025}*250 = 11250

The 90th percentile for the distribution of the total contributions

This is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. Then

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

1.28 = \frac{X - 6328125}{11250}

X - 6328125 = 1.28*11250

X = 6342525

The 90th percentile for the distribution of the total contributions is $6,342,525.

3 0
1 year ago
In the figure above, two line segments in the x-y plane form a right triangle with the x-axis. What is the area of the triangle
Juli2301 [7.4K]
A = (1/2) * b * h
The base is 10; the x-value going from 0 to 10.
The perpendicular height is 2; the y-value going from 0 to 2
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2 years ago
Which diagram represents a fraction equivalent to 75%?
Lapatulllka [165]

Answer:

Honestly i think i know which diagrams your talking about but if it there is an option that has 20 squares and only 15 are shaded then thats the one u choose I'm thinking that option is c

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7 0
2 years ago
Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

5 0
1 year ago
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