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2 years ago

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Nakita bought items at a grocery store.  She bought 2 boxes of crackers for $3.50 each.  She used a coupon for $0.80 off the p

rice of each box of crackers.  She bought a jar of peanut butter for $4.85.  She bought a package of juice boxes for $2.40.  She used a coupon for $3.00 off the total price of the items she bought. This expression can be used to determine the price of the items Nakita bought. [2(3.50 − 0.80) + 4.85 + 2.40] − 3.00 What is the price of the items Nakita bought? A. $6.95 B. $10.45 C. $9.65 D. $12.65

1 answer:

olga nikolaevna [1]2 years ago

4 0

Answer: $12.65

Step-by-step explanation: sikeeeeeeeeeeee

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Please help with 1 , and 2 I really don’t understand this!

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1. B
2. A

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1 year ago

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Chang knows one side of a triangle is 13 cm. Which set of two sides is possible for the lengths of the other two

Vinvika [58]

Answer:

8cm and 9cm

Step-by-step explanation:

8cm and 9cm would be the two possible lengths for the triangle because for the triangle to work you need the other two other sides when added together needs to be greater then 13. Furthermore 5 + 8 = 13 and as 13 is the same length as the one side that is given, a triangle couldn't possibly be formed as it would just be a straight line. Moreover, this is the same with 6 + 7 which is also 13. 7 + 2 is 9 and because 9<13 a triangle couldn't possibly be formed. Finally, 8 + 9 which equals to 17 are the only two possible lengths for the triangle as 17 > 12.

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2 years ago

LekaFEV [45]

6 bills in the deposit were $10 bills.

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

The total number of bills he gave the teller = 24 bills

x + y = 24 ------------(1)

Let,

x be the number of $5 bills.

y be the number of $10 bills.

The total bank deposit made using $5 and $10 bills are $150.

<u>He deposits $5 bills and $10 bills. So, the equation can be framed as : </u>

5x + 10y = 150 --------(2)

The question is asked for how many $10 bills were in the deposit,

Here 'y' is the number of $10 bills.

<u>To solve for y value : </u>

The eq(1) must be multiplied by 5 and subtract eq(2) from eq(1),

5x + 5y = 120

-(<u>5x + 10y = 150</u>)

<u>-5y = -30 </u>

⇒ y = 30/5

⇒ y = 6

Therefore, there were 6 bills in the deposit that are $10 bills.

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2 years ago

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Answer:

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How many randomly chosen guests should I invite to my party so that the probability of having a guest with the same birthday as

dimulka [17.4K]

Answer:

At least 401 chosen guests

Step-by-step explanation:

Represent those with same birth day with Same

So:

$P(Same) > \frac{2}{3}$

Represent number of people with n

There are 365 days in a year and 364 days out of these days is not your birthday.

So, there the probability that n people do not share your birthday is $(\frac{364}{365})^n$

i.e.

$P(none) = (\frac{364}{365})^n$

Solving further, we have that:

$P(same) + P(none) = 1$

$P(same) + (\frac{364}{365})^n = 1$

$P(same) = 1 - (\frac{364}{365})^n$

We're calculating the probability that $P(Same) > \frac{2}{3}$ .

So, we have:

$1 - (\frac{364}{365})^n > \frac{2}{3}$

Collect Like Terms

$1 - \frac{2}{3}> (\frac{364}{365})^n$

$\frac{3-2}{3}> (\frac{364}{365})^n$

$\frac{1}{3}> (\frac{364}{365})^n$

$3^{-1} > (\frac{364}{365})^n$

Take natural logarithm (ln) of both sides

$ln(3^{-1}) > ln((\frac{364}{365})^n)$

$-ln\ 3 > n\ ln\ (\frac{364}{365})$

Apply laws of logarithm

$-ln\ 3 > n\ (ln(364) - ln(365))$

Multiply through by -1

$ln\ 3 < n\ (ln(365) - ln(364))$

Solve for n

$\frac{ln\ 3}{(ln(365) - ln(364))} < n$

Reorder

$n > \frac{ln\ 3}{(ln(365) - ln(364))}$

$n > \frac{1.09861228867}{(5.89989735358 - 5.89715386764)}$

$n> \frac{1.09861228867}{0.00274348594}$

$n > 400.443928891$

$n > 400$ (approximated)

This implies that n = 401, 402, 403 .....

i.e

$n \geq 401$

So, at least 401 people has to be invited

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