66,649 is 70% of what number

in the school cafeteria students chose their lunch from 3 sandwiches, 3 soups, 4 salads, and 2 drinks. how many different lunche

Vesna [10]

Answer:

Answer:

3

×

3

×

4

×

2

=

72

Explanation:

Let's look at the 3 sandwiches and 3 soups first and then expand the calculation. There are 9 ways I can have one of the sandwiches and 1 of the soups:

⎛

⎜

⎝

0

Soup 1

Soup 2

Soup 3

Sandwich 1

1

2

3

Sandwich 2

4

5

6

Sandwich 3

7

8

9

⎞

⎟

⎠

And so we can see that we multiply the number of sandwiches and the number of soups to get the total number of ways to get one of each.

The same works for more categories of choices, and so we multiply the 3 sandwiches, the 3 soups, 4 salads, and 2 drinks to get:

3

×

3

×

4

×

2

=

72

7 0

3 years ago

Evaluate the expression 8 power of 2+3.8-10 times 2

Nitella [24]

Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

-----------------------------------------------------------------------

$8^2+3.8-10\cdot2=64+3.8-20=67.8-20=47.8$

6 0

3 years ago

Please help i dont understand this at all

tamaranim1 [39]

I have no idea but I googled and I got 0.6 I hope this helps!

5 0

4 years ago

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 Y is defined as the number of consecutive time intervals in which the water supply remains below a critical value y₀.

The random variable Y follows a Geometric distribution with parameter p = 0.409.

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 Y as follows:

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

Compute the standard deviation of the random variable Y 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

4 years ago

A biologist estimated that the current population of ladybugs in the region was 8.3 x 106. He predicts that a swarm of

stiks02 [169]

Answer:

iId

Step-by-step explanation:

6 0

3 years ago