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german
2 years ago
6

A local store is advertising a shirt on sale for 40% off the original price. The sale price is $21.00. What was the original pri

ce for the shirt?
Mathematics
1 answer:
pishuonlain [190]2 years ago
8 0

Answer:

  $35

Step-by-step explanation:

If $21 is 60% of the original price, then the original price is ...

  21 = 0.60p

  21/0.60 = p = 35

The original price is $35.00.

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10. Fred has to read a book for a test in 2 weeks. If he reads the same number of pages per day and the book has 350 pages, how
Angelina_Jolie [31]

Answer:

25 pages per day

Step-by-step explanation:

Fred has 14 days to finish 350 pages.

To find out how many pages he needs to read each day, we divide 350 by 14, to get 25.

5 0
1 year ago
Read 2 more answers
4+8+12=3x8 the next equation is
MAXImum [283]

Answer:

True 24=24

Step-by-step explanation:

4+8+12=3x8

12+12=24

24=24

7 0
3 years ago
Which statement is true about every rectangular pyramid?
Ratling [72]

Answer:

At least two of the lateral faces are congruent.

Step-by-step explanation:

Because the words "at least" are used this means that at a minimum 2 of the faces are congruent. Because the base is in the shape of a triangle, this will definitely be true.

hope this helps:)

3 0
2 years ago
Read 2 more answers
Pls I need help on this problem
Liono4ka [1.6K]

Answer:

x = 64

Step-by-step explanation:

they give you two angles 56 and 60

add them and subtract from 180

180 - 116 = 64

7 0
3 years ago
A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective
Inessa05 [86]

Answer:

(a) P(X \leq 20) = 0.9319

(b) Expected number of defective light bulbs = 15

Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....

where, n = number of samples taken = 150

            r = number of success

           p = probability of success which in our question is % of bulbs that

                  are defective, i.e. 10%

<em>Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.</em>

So, Let X = No. of defective bulbs in a box

<u>Mean of X</u>, \mu = n \times p = 150 \times 0.10 = 15

<u>Standard deviation of X</u>, \sigma = \sqrt{np(1-p)} = \sqrt{150 \times 0.10 \times (1-0.10)} = 3.7

So, X ~ N(\mu = 15, \sigma^{2} = 3.7^{2})

Now, the z score probability distribution is given by;

                Z = \frac{X-\mu}{\sigma} ~ N(0,1)

(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X \leq 20) = P(X < 20.5)  ---- using continuity correction

    P(X < 20.5) = P( \frac{X-\mu}{\sigma} < \frac{20.5-15}{3.7} ) = P(Z < 1.49) = 0.9319

(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = n \times p = 150 \times 0.10 = 15.

                                           

5 0
3 years ago
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