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vladimir2022 [97]

3 years ago

6

An astronomy club wants to buy six new eyepieces for their telescope. Four of the lenses cost $53.98 each. Two lenses cost $21 e

ach. What is the total cost?

1 answer:

Igoryamba3 years ago

7 0

The answer is 257.92 dollars. Comment if you need more help :)

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Someone please helppp.....

Alecsey [184]

I believe that the answer would be B) y = x + 19

3 0

3 years ago

Read 2 more answers

Which of the following is not a rational number?

Illusion [34]

D because a rational number can be converted into a fraction and the square root of 5 cannot be covert into a fraction because it equals 2.2

4 0

3 years ago

Suppose that X1 and X2 are independent random variables each with a mean μ and a variance σ^2. Compute the mean and variance of

Deffense [45]

Mean:

E[Y] = E[3X₁ + X₂]

E[Y] = 3 E[X₁] + E[X₂]

E[Y] = 3µ + µ

E[Y] = 4µ

Variance:

Var[Y] = Var[3X₁ + X₂]

Var[Y] = 3² Var[X₁] + 2 Covar[X₁, X₂] + 1² Var[X₂]

(the covariance is 0 since X₁ and X₂ are independent)

Var[Y] = 9 Var[X₁] + Var[X₂]

Var[Y] = 9σ² + σ²

Var[Y] = 10σ²

5 0

3 years ago

Find the missing measure in a right triangle if a= 9 ft and b = 12 ft.

Lena [83]

Answer:

c = 15 ft

Step-by-step explanation:

a^2 + b^2 = c^2

9^2 + 12^2 = c^2

c^2 = 225

c = √225

<u>c = 15 ft</u>

5 0

3 years ago

Use sigma notation to represent the sum of the first seven terms of the following sequence -4,-6,-8.....

lina2011 [118]

Answer:Answer:

$\sum\left {{7} \atop {1}} \right -n(3+n)$

Step-by-step explanation:

Given the sequence -4,-6,-8..., in order to get sigma notation to represent the sum of the first seven terms of the sequence, we need to first calculate the sum of the first seven terms of the sequence as shown;

The sum of an arithmetic series is expressed as $S_n = \frac{n}{2}[2a+(n-1)d]$

n is the number of terms

a is the first term of the sequence

d is the common difference

Given parameters

n = 7, a = -4 and d = -6-(-4) = -8-(-6) = -2

Required

Sum of the first seven terms of the sequence

$S_7 = \frac{7}{2}[2(-4)+(7-1)(-2)]\\\\S_7 = \frac{7}{2}[-8+(6)(-2)]\\\\S_7 = \frac{7}{2}[-8-12]\\\\\\S_7 = \frac{7}{2} * -20\\\\S_7 = -70$

The sum of the nth term of the sequence will be;

$S_n = \frac{n}{2}[2(-4)+(n-1)(-2)]\\\\S_n = \frac{n}{2}[-8+(-2n+2)]\\\\S_n = \frac{n}{2}[-6-2n]\\\\S_n = \frac{-6n}{2} - \frac{2n^2}{2}\\S_n = -3n-n^2\\\\S_n = -n(3+n)$

The sigma notation will be expressed as $\sum\left {{7} \atop {1}} \right -n(3+n)$ . <em>The limit ranges from 1 to 7 since we are to find the sum of the first seven terms of the series.</em>

3 0

3 years ago