Ask question

Ask question

All categories

2 years ago

13

Margo purchase teas from a warehouse and marks up the price 20%. When teas got unsold for two mouths, she marked the price down

for 20%. She says that she is breathing even, that is, she is getting the sam price for the teas that she paid for it. Is she correct? Explain.

1 answer:

sladkih [1.3K]2 years ago

3 0

No because you are going to get more money off.You have to make a proposition to help solve this.

You might be interested in

Using the figure below, find the area of the circle.

jeyben [28]

Refers to the attachment

6 0

2 years ago

Is (5,1) a solution to the equation 2x-y=11? Explain

grigory [225]

This is simple. Coordinate form is (x,y) so you would just plug in your values.

2(5)-(1)=11

10-1 does not equal 11, so no it is not a solution.

7 0

3 years ago

Read 2 more answers

A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as

Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

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

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

$P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888$

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

$=1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776$

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

$={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453$

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0

3 years ago

LABEL THE TRIANGLE!! PLEASE HELP!!

Anettt [7]

Answer:

Determination of HYP,OPP,ADJ with respect to x.

<u>Opposite</u><u> </u><u>side</u><u> </u><u>of</u><u> </u><u>right</u><u> </u><u>angle</u><u>:</u><u>Hypotenuse</u><u>:</u><u> </u><u>AC</u>

<u>Opposite</u><u> </u><u>side</u><u> </u><u>of</u><u> </u><u>given</u><u> </u><u>angle</u><u>:</u><u> </u><u>Opp</u><u>:</u><u>BC</u>

<u>remaining</u><u> </u><u>side</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>triangle</u><u>:</u><u> </u><u>Adjacent</u><u>:</u><u>AB</u><u>.</u>

8 0

2 years ago

Which of the following shows the number 7.12 x 10^-3 written in standard form? Answer choices : 0.00712 : 0.0712 : 0.712 : 7,120

svet-max [94.6K]

<span>written in standard form
</span><span>
7.12 x 10^-3 = 0.00712

hope it helps</span>

7 0

3 years ago