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vfiekz [6]
3 years ago
6

I need help please help me

Mathematics
1 answer:
Eduardwww [97]3 years ago
3 0

Answer:

Option 2

Step-by-step explanation:

Cost of book= 6.50

Shipping= 4.99

Total= 82.99

Equation:

6.50x+4.99= 82.99

Option 2 is correct

You might be interested in
A total of 815 tickets were sold for the school play. They were either adult tickets or student tickets. There were 65 more stud
kvv77 [185]

Answer:

<u>375 Adult Tickets.</u>

Step-by-step explanation:

Here, we can simply set up an equation using variable <em>x </em>in place of the unknown student/adult tickets.

x = # of <u>adult</u> tickets sold

x + 65 = # of <u>student</u> tickets sold.

1) x + x + 65 = 815 (set both ticket amounts equal to the total)

2) 2x + 65 = 815 (added common variables together)

3) 2x = 750 (negated the +65, subtracted it from both sides)

4) x = 375 (divided both sides by 2)

5) 815 - 375 = 440 (subtracted the x from the total number of <u>adult</u> tickets, to recieve the amount of <u>childrens</u>' tickets.

Therefore,

Since there were fewer adult tickets sold (-65), 375 is the number of adult tickets, and 440 is the number of student tickets.

5 0
3 years ago
What is 3a squared add 7 a
vfiekz [6]
The answer is:  " 9a² + 7a " .
_________________________________
Note:    (3a)² + 7a = (3a)*(3a) + 7a = 9a² + 7a .
_________________________________
3 0
3 years ago
Convert to standard form y=2(x+3)(x-6)
Nataly_w [17]

Answer:

the answer is 2x²-6x-36

3 0
3 years ago
manufacturing company produces digital cameras and claim that their products maybe 3% defective. A video company, when purchasin
alexdok [17]

Answer:

P(X>17) = 0.979

Step-by-step explanation:

Probability that a camera is defective, p = 3% = 3/100 = 0.03

20 cameras were randomly selected.i.e sample size, n = 20

Probability that a camera is working, q = 1 - p = 1 - 0.03 = 0.97

Probability that more than 17 cameras are working P ( X > 17)

This is a binomial distribution P(X = r) nCr q^{r} p^{n-r}

nCr = \frac{n!}{(n-r)!r!}

P(X>17) = P(X=18) + P(X=19) + P(X=20)

P(X=18) = 20C18 * 0.97^{18} * 0.03^{20-18}

P(X=18) = 20C18 * 0.97^{18} * 0.03^{2}

P(X=18) = 0.0988

P(X=19) = 20C19 * 0.97^{19} * 0.03^{20-19}

P(X=19) = 20C19 * 0.97^{19} * 0.03^{1}

P(X=19) = 0.3364

P(X=20) = 20C20 * 0.97^{20} * 0.03^{20-20}

P(X=20) = 20C20 * 0.97^{20} * 0.03^{0}

P(X=20) = 0.5438

P(X>17) = 0.0988 + 0.3364 + 0.5438

P(X>17) = 0.979

The probability that there are more than 17 working cameras should be 0.979 for the company to accept the whole batch

6 0
3 years ago
Help!!i will give BRAINLIEST to whoever answers
Gnom [1K]

Answer:

-11.167 is the answrr

Step-by-step explanation:

5 0
3 years ago
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