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Arlecino [84]
3 years ago
5

(-18 divided by - 2) + ( 28 divided by 7)

Mathematics
1 answer:
andrew11 [14]3 years ago
6 0

Answer:

-18 divided by -2 should be -9. You add 18 divided by 7 ,which is 4, and this should make it -5.

Step-by-step explanation:

Hope this helps!

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A trail is 2 miles long. There is a marker every 1/8 mile. How many markers are there?
attashe74 [19]

Answer:

16

Step-by-step explanation:

Number of Markers =

2 \div  \frac{1}{8}  \\  = 2 \times 8 \\  = 16

8 0
3 years ago
Reduced ratio 27 to 9
Sidana [21]
3:1 is your answer :)
3 0
3 years ago
A toy store’s percent of markup is 30%. A model train costs the store $50. Find the markup.
Over [174]

Answer:  $15

Step-by-step explanation:

To find the markup price just multiply 30% by the amount.

30% * 50  

0.3 * 50 = 15

4 0
3 years ago
Read 2 more answers
According to the article "Characterizing the Severity and Risk of Drought in the Poudre River, Colorado" (J. of Water Res. Plann
mihalych1998 [28]

Answer:

(a) P (Y = 3) = 0.0844, P (Y ≤ 3) = 0.8780

(b) The probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

Step-by-step explanation:

The random variable <em>Y</em> is defined as the number of consecutive time intervals in which the water supply remains below a critical value <em>y₀</em>.

The random variable <em>Y</em> follows a Geometric distribution with parameter <em>p</em> = 0.409<em>.</em>

The probability mass function of a Geometric distribution is:

P(Y=y)=(1-p)^{y}p;\ y=0,12...

(a)

Compute the probability that a drought lasts exactly 3 intervals as follows:

P(Y=3)=(1-0.409)^{3}\times 0.409=0.0844279\approx0.0844

Thus, the probability that a drought lasts exactly 3 intervals is 0.0844.

Compute the probability that a drought lasts at most 3 intervals as follows:

P (Y ≤ 3) =  P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3)

              =(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409+(1-0.409)^{2}\times 0.409\\+(1-0.409)^{3}\times 0.409\\=0.409+0.2417+0.1429+0.0844\\=0.8780

Thus, the probability that a drought lasts at most 3 intervals is 0.8780.

(b)

Compute the mean of the random variable <em>Y</em> as follows:

\mu=\frac{1-p}{p}=\frac{1-0.409}{0.409}=1.445

Compute the standard deviation of the random variable <em>Y</em> as follows:

\sigma=\sqrt{\frac{1-p}{p^{2}}}=\sqrt{\frac{1-0.409}{(0.409)^{2}}}=1.88

The probability that the length of a drought exceeds its mean value by at least one standard deviation is:

P (Y ≥ μ + σ) = P (Y ≥ 1.445 + 1.88)

                    = P (Y ≥ 3.325)

                    = P (Y ≥ 3)

                    = 1 - P (Y < 3)

                    = 1 - P (X = 0) - P (X = 1) - P (X = 2)

                    =1-[(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409\\+(1-0.409)^{2}\times 0.409]\\=1-[0.409+0.2417+0.1429]\\=0.2064

Thus, the probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

6 0
3 years ago
X2+6x-16) divided (x-2)
Nady [450]

Answer:

let's hope for the best ....XD

5 0
3 years ago
Read 2 more answers
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