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Lapatulllka [165]
3 years ago
12

4.) You plan to start making a series of deposits of $3,000 a year for 6 years. How much will you have after 6 years if your acc

ount has an annual interest rate of 6%?​
Mathematics
1 answer:
stich3 [128]3 years ago
5 0

Answer:16,920

Step-by-step explanation:

first convert 6% to a decimal by moving the decimal point two times to the left. then multiply 3,000 by 0.06 which will equal 180 for one year. you then multiply 180 by 6 and get 1080. after you multiply 3000 by 6 which is 18000. subtract 1,080 from 18000 and you’ll get 16920

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A company makes baseball cards in packs of 10, boxes of 100, and cases of 1,000. A store orders a total of 15,470 cards. How can
Wewaii [24]

A company makes baseball cards in packs of 10, boxes of 100, and cases of 1,000. A store orders a total of 15,470 cards. How can the company fill the order in the exact amount?

Answer: The company can fill the order in exact numbers as:

15 cases, 4 boxes, 7 packs

15 cases contain 15 \times 1000=15000 cards

4 boxes contain 4 \times 100 =400 cards

7 packs contain 7 \times 10 = 70 cards

Therefore, the total cards are:

15000 + 400 + 70 = 15,470

4 0
4 years ago
On the subway 8 out of 11 people are carrying a briefcase. Based on this information, if there are 700 people on the subway, the
leva [86]

Answer: 127

Step-by-step explanation:

8 0
3 years ago
From a shipment of 65 transistors, 6 of which are defective, a sample of 5 transistors is selected at random.
Vladimir79 [104]

Answer:

a) 8259888

b) 34220

c) 45057474

Step-by-step explanation:

Given,

The total number of transistor = 65,

In which, the defective transistor = 6,

So, the number of non defective transistor = 65 - 6 = 59,

Since, out of these transistor 5 are selected,

a) Thus, the number of ways = the total possible combination of 5 transistors = {65}C_ 5

=\frac{65!}{(65-5)!5!}

=8259888

b) The number of samples that contains exactly 3 defective transistors = the possible combination of exactly 3 defective transistors = {6}C_3\times {59}C_2

=\frac{6!}{(6-3)!3!}\times \frac{59!}{(59-2)!\times 2!}

=20\times 1711

=34220

c) The number of sample without any defective transistor = The possible combination of 0 defective transistor = ^6C_0\times ^{59}C_5

=1\times 45057474

=45057474

5 0
3 years ago
Factor completely 2x3 + 6x2 + 2x + 6.
katovenus [111]

The answer is 2(x + 3)(x^2 + 1).

Steps:

2x^3 + 6x^2 + 2x + 6

2x^3 + 2x + 6x^2 + 6

2x(x^2 + 1) + 6(x^2 + 1)

(2x + 6)(x^2 + 1)

2(x + 3)(x^2 + 1)

4 0
3 years ago
Read 2 more answers
One of the legs of a right triangle is twice as long as the other, and the perimeter of the triangle is 28. Find the length of t
Nutka1998 [239]

1) If we let the lengths of the legs be x and 2x, then by the Pythagorean theorem,

x+2x+\sqrt{x^{2}+(2x)^{2}}=28\\3x+x\sqrt{5}=28\\x(3+\sqrt{5})=28\\x=\frac{28}{3+\sqrt{5}} \cdot \frac{3-\sqrt{5}}{3-\sqrt{5}}=21-7\sqrt{5}

This means that the length of the hypotenuse is:

\sqrt{5}(21-7\sqrt{5})=\boxed{21\sqrt{5}-35}

2) (Diagram attached) We know that

\cos 51^{\circ}=\frac{x}{58}\\x=58 \cos 51^{\circ} \approx \boxed{36.5}  (in cm)

3) From the Pythagorean identity,

\sin^{2}\left (\arccos \frac{5}{13} \right)+\cos^{2} \left (\arccos \frac{5}{13} \right)=1\\\left(\frac{5}{13} \right)^{2}+\sin^{2} \left (\arccos \frac{5}{13} \right)=1\\\frac{25}{169}+\sin^{2} \left (\arccos \frac{5}{13} \right)=1\\\sin^{2} \left (\arccos \frac{5}{13} \right)=\frac{144}{169}\\\sin \left (\arccos \frac{5}{13} \right)=\boxed{\frac{12}{13}}

4) \frac{\pi}{6}

7 0
2 years ago
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