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dem82 [27]
2 years ago
5

Mario invested $6,000 in an account that pays 5% annual interest compounded annually. Using the formula A = P(1 + r)t, what is t

he approximate value of the account after 2.5 years?
a. $6,075
b. $6,118
c. $6,456
d. $6,778
Mathematics
1 answer:
jok3333 [9.3K]2 years ago
6 0
Given:
Principal = 6,000
interest rate = 5%
term = 2.5 years

A = P (1+r)^t
A = 6,000 (1 + 0.05)².⁵
A = 6,000 (1.05)².⁵
A = 6,000 (1.1297)
A = 6,778.20 Choice D.

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The list shows numbers in order from least to greatest. Negative 34, negative 2 and three-fourths, blank, negative 1.5, 0, 2.3,
blsea [12.9K]

Options

  • -2.5
  • -2
  • -1\dfrac{9}{10}
  • -1

Answer:

(B) -2

Step-by-step explanation:

Given the list

-34.-2\dfrac{3}{4} ,x, -1.5,0,2.3,10,25\dfrac15

This list has been arranged in order from the least to the greatest.

We are required to find an Integer that can be inserted on the blank line (I have used x) in the list.

An integer is defined as a positive or negative whole number.

Out of the given options, only -1 and -2 are integers. However:

  • -1 is greater than -1.5

Therefore, it cannot be our result.

The correct answer is -2.

3 0
2 years ago
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An airliner carries 150 passengers and has doors with a height of 78 in. Heights of men are normally distributed with a mean of
iogann1982 [59]

Answer:

Step-by-step explanation:

x, height of men is N(69, 2.8)

Sample size n =150

Hence sample std dev = \frac{2.8}{\sqrt{150} } =0.229

Hence Z score = \frac{x-69}{0.229}

A) Prob that a random man from 150  can fit without bending

= P(X<78) = P(Z<3.214)=1.0000

B) n =75

Sample std dev = \frac{2.8}{\sqrt{75} } =0.3233

P(X bar <72) = P(Z<9.28) = 1.00

C) Prob of B is more relevent because average male passengers  would be more relevant than a single person

(D) The probability from part (b) is more relevant because it shows the proportion of flights where the mean height of the male passengers will be less than the door height.

3 0
2 years ago
Solving exponential growth and decay problems
Paul [167]
See tha attached pdf file and let me know how useful this answer is.
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6 0
2 years ago
RQ is tangent to Circle S. Use the diagram below to find the measure of QRS. mQRS &lt; 90 mQRS &gt; 90 mQRS = 90 not enough info
Masteriza [31]
Given that you did not include the diagram showing the circle, the tangent line and the points Q, R, and S, I am going to give you the explanation to answer the question.

1) The tangent lines to a circle form a 90° angle with the radius at the point of intersection.

2) Therefore, if the point of intersection of the tangent line and the circle is named R, and the points S and Q are one the center of the circle and the other is on the line RQ, then you know that the segment SR is a radius and the line RQ is the tangent, which means that they are perpendicular, i.e. the angle QRS is measures 90°.

In this case the answer is m angle QRS = 90°.

3) Otherwise the angle is different to 90° and you need to observe the figure to conclude whether it is greater than 90°, less than 90° or there is not enough information.
7 0
2 years ago
Factor the expression given below. Write each factor as a polynomial in
kherson [118]

The factorization of the expression of 43x³ + 216y³ is

(7x + 6y)(49x² - 42xy + 36y²)

Step-by-step explanation:

The sum of two cubes has two factors:

1. The first factor is \sqrt[3]{1st} + \sqrt[3]{2nd}

2. The second factor is ( \sqrt[3]{1st} )² - ( \sqrt[3]{1st} ) ( \sqrt[3]{2nd} ) + ( \sqrt[3]{2nd} )²

Ex: The expression a³ + b³ is the sum of 2 cubes

The factorization of a³ + b³ is (a + b)(a² - ab + b²)

∵ The expression is 343x³ + 216y³

∵ \sqrt[3]{343x^{3}} = 7x

∵ \sqrt[3]{216y^{3}} = 6y

∴ The first factor is (7x + 6y)

∵ (7x)² = 49x²

∵ (7x)(6y) = 42xy

∵ (6y)² = 36y²

∴ The second factor is (49x² - 42xy + 36y²)

∴ The factorization of 43x³ + 216y³ is (7x + 6y)(49x² - 42xy + 36y²)

The factorization of the expression of 43x³ + 216y³ is

(7x + 6y)(49x² - 42xy + 36y²)

Learn more:

You can learn more about factors in brainly.com/question/10771256

#LearnwithBrainly

6 0
2 years ago
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