MATHEMATICS HIGH SCHOOL

Find the first,fourth, and eighth terms of the sequence. A(n)=-5*2^n-1

Answers

Answer 1

Answer:

Answer: $A(1)=-5$

$A(4)=-40$

$A(8)=-640$

Step-by-step explanation:

Given sequence : $A(n)=-5*2^(n-1)$

To find the first term, we need to put n=1 in the above sequence , we get

$A(1)=-5*2^(1-1)=-5*2^0=-5$

To find the fourth term, we need to put n=4, we get

$A(4)=-5*2^(4-1)=-5*2^3=-5*8=-40$

To find the eighth term, we need to put n=8, we get

$A(8)=-5*2^(8-1)=-5*2^7=-5*128=-640$

Answer 2

Answer:

Answer:

$A(1)=-5*2^1-1=-11$

$A(1)=-5*2^4-1=-81$

$A(1)=-5*2^8-1=-1281$

Step-by-step explanation:

Given : $A(n)=-5*2^n-1$

To find : $A(1), A(4), A(18)$

Solution:

Put n=1, $A(1)=-5*2^1-1$

$A(1)=-10-1=-11$

Put n=4, $A(1)=-5*2^4-1$

$A(1)=-80-1=-81$

Put n=8, $A(1)=-5*2^8-1=$

$A(1)=-1280-1=-1281$

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The weights for a population of North American raccoons have a bell-shaped frequency curve with a mean of about 12 pounds and a standard deviation of about 2.5 pounds. About 95% of the weights for individual raccoons in this population fall between what two values?

Answers

Answer:

Step-by-step explanation:

Given that the weights for a population of North American raccoons have a bell-shaped frequency curve with a mean of about 12 pounds and a standard deviation of about 2.5 pounds

Since bell shaped is given we can assume that this follows a normal distribution symmetrical about the mean.

To find 95% two values we can use Z critical value for 95%

i.e. ±1.96

95% will lie between 12±1.96(2.5)

= 12±4.9

=(7.1, 16.9)

A school is having a canned food drive. Each class is challenged to collect 200 cans. If there are c classes in the school, which equation could be used to find the total number of cans, T, the school will collect if each class meets the challenge? A. T= c/200 B. T=200/c

C. 200c

D. 200+c

Answers

The answer would be C. 200c. This is because in order to find the answer you multiply the number of cans (200) by the number of classes (c) to get the total (t) for the school. Therefore, your equation would be: T= 200c

What is the sum of the first 39 positive odd numbers?

Answers

1-39 ODD:
1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39
Now you'd add them all up.
1+3+5+7+9 = 25
11+13+15+17+19 = 75
21+23+25+27+29 = 125
31+33+35+37+39 = 175
Add the totals:
25+75+125+175
The answer would be 400.

Final answer:

The sum of the first 39 positive odd numbers is 1521.

Explanation:

To find the sum of the first 39 positive odd numbers, we can use the formula for the sum of an arithmetic series. The formula is: S = (n/2)(a + l), where S is the sum, n is the number of terms, a is the first term, and l is the last term. In this case, n = 39, a = 1, and l = 77. Plugging these values into the formula, we get:

S = (39/2)(1 + 77)
S = (39/2)(78)
S = 39 * 39
S = 1521

Therefore, the sum of the first 39 positive odd numbers is 1521.

Learn more about Sum of odd numbers here:

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If x is an integer, then is -x positive?

Answers

Answer:

If x is an integer, then for values of x ≤ 0 would -x be positive.

General Formulas and Concepts:

Math

Integers

Step-by-step explanation:

We know that integers comprise of the number line from -∞ to ∞. We can have numbers like -3, -2, -1, 0, 1, 2 ,3.

If we say that x is an integer, and that -x must be positive, then that means the integer x must be negative, because a negative times a negative is a positive.

∴ x can only be negative integers, thus giving us x ≤ 0.

In how many ways can 50 cards be chosen from a standard deck of 52 cards using permutations

Answers

In 26*51! ways we can pick 50 cards from a deck of 52 cards using permutation.

What is Permutation?

Permutation is mathematical technique to get possibiltiy in arrangement with replacement.

We can choose or pick r objects from n objects using permutation in $^nP_r$ ways.

$^nP_r=(n!)/((n-r)!)$

n! = n*(m-1)*......*3*2*1

We can write n! = n(n-1)!

Total number of matters or cards = 52

We have to pick 50 cards using permutation or replacement.

The required number of ways $=^(52)P_(50)=(52!)/((52-50)!)=(52*51!)/(2!)=26*51!$

Hence we can choose 50 cards from deck of 52 cards in 6*51! ways.

Learn more about Permutation here -

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Permutation of 50 is 50! or 50*49*48*47*46.... and so on. I doubt they actually want you to find out what 50*49*48... is, so I'd just put 50!

Liz owns stock in Nar Heating/Cooling and Cilla Shipping. She owns 120 more shares of Cilla Shipping than she does of Nar Heating/Cooling. Nar Heating/Cooling pays a yearly dividend of $5.78 per share, and Cilla Shipping pays a yearly dividend of $4.15 per share. If Liz receives $1630.02 in dividends annually, how many shares of Nar Heating/Cooling does she own?

Answers

The number of shares of Nar Heating/Cooling that Liz has is 114. Let x be the number of shares in Nar Heating/Cooling and y the number of shares in Cilla Shipping. Liz owns 120 more shares of Cilla Shipping than she does of Nar Heating/Cooling: y = x + 120. Liz receives $1630.02 in dividends annually: 5.78x + 4.15y = 1630.02. This is the system of two equations: y = x + 120 and 5.78x + 4.15y = 1630.02. Substitute y from the first equation into the second one: 5.78x + 4.15(x + 120) = 1630.02. 5.78x + 4.15x + 498 = 1630.02. 9.93x + 498 = 1630.02. 9.93x = 1630.02 - 498. 9.93x = 1132.02. x = 1132.02/9.93 = 114. So, the number of shares in Nar Heating/Cooling is 114.

Answer:

Edge answer is B) 114 shares

Step-by-step explanation:

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