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2 years ago

The number of elementary and high school students who joined the math club may be expressed as 3⁴ + 1. Evaluate 3⁴ + 1.

Mathematics

2 answers:

victus00 [196]2 years ago

6 0

Answer:

3^4 + 1 = 82

Step-by-step explanation:

3^4 = 81

81 + 1 = 82

kogti [31]2 years ago

5 0

Answer:

Step-by-step explanation:

3⁴ + 1

3 × 3 × 3 × 3 + 1

9 × 9 + 1

81 + 1

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Paula wants to save at least $800 for a trip. Which inequality shows the least amount she must save each month for 9 months to a

ElenaW [278]

Answer:

Step-by-step explanation:

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3 years ago

ASAP PLEASE HELP! THANK YOUU

mote1985 [20]

The answer to the question is c

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3 years ago

4x^2-21x-18=0<br> Find the discriminate

SpyIntel [72]

a = 4, b = -21, c = -18

to keep from getting "mixed up", evaluate the discriminant first ...

b<sup>2</sup> - 4ac = (-21)<sup>2</sup> - 4(4)(-18) = 729

sqrt(729) = 27

x = (21 +/- 27)/8

x = -3/4, x = 6

since the discriminant is a perfect square, the original quadratic will factor ...

4x<sup>2</sup> - 21x - 18 = 0

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6 0

3 years ago

Read 2 more answers

Suppose you just received a shipment of nine televisions. Three of the televisions are defective. If two televisions are randoml

Temka [501]

<h2>Answer:</h2>

The probability that both televisions work is: 0.42

The probability at least one of the two televisions does not work is:

0.5833

<h2>Step-by-step explanation:</h2>

There are a total of 9 televisions.

It is given that:

Three of the televisions are defective.

This means that the number of televisions which are non-defective are:

9-3=6

The probability that both televisions work is calculated by:

$=\dfrac{6_C_2}{9_C_2}$

( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.

and the total outcome is the selection of 2 televisions from a total of 9 televisions)

Hence, we get:

$=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42$

The probability at least one of the two televisions does not work:

Is equal to the probability that one does not work+probability both do not work.

Probability one does not work is calculated by:

$=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5$

and the probability both do not work is:

$=\dfrac{3_C_2}{9_C_2}\\\\\\=\dfrac{1}{12}\\\\\\=0.0833$

Hence, Probability that atleast does not work is:

0.5+0.0833=0.5833

4 0

4 years ago

Alicia has five times as many dimes as she has quarters. Combined the dimes and quarters total to $9.75. How many dimes does Ali

olasank [31]

Five dimes and one quarter = .75.
9.75/.75=3.25*4/1=13
there are 13 "5 dimes and 1 quarter"s
so 65 dimes

3 0

3 years ago