What is the sum?

agasfer [191]

There's always a decimal at the end of a whole number. here it's behind the last zero. move that decimal so that the non-zero digit is less than 10 but more than 1.
4.73, the decimal was moved 4 places to the left. so the scientific notation is
4.72X10^4

3 0

3 years ago

anonymous 2 years ago If you were to solve the following system by substitution, what would be the best variable to solve for an

Ainat [17]

I'm sorry that I'm not in calculus, but hopefully I can help.
Personally, I would choose: D.) x; in the second equation ;
because it has easier numbers to figure out, you'll get it eventually...

7 0

3 years ago

Y is a random variable that is distributed N(-16, 1.21). Find k such that Prob(-15.043 < Y ≤ k) = 0.1546. (Round your answer

IgorLugansk [536]

Transform Y to Z, which is distributed N(0, 1), using the formula

Y = µ + σZ

where µ = -16 and σ = 1.21.

Pr[-15.043 < Y ≤ k] = 0.1546

Pr[(-15.043 + 16)/1.21 < (Y + 16)/1.21 ≤ (k + 16)/1.21] = 0.1546

Pr[0.791 < Z ≤ (k + 16)/1.21] ≈ 0.1546

Pr[Z ≤ (k + 16)/1.21] - Pr[Z < 0.791] = 0.1546

Pr[Z ≤ (k + 16)/1.21] = 0.1546 + Pr[Z < 0.791]

Pr[Z ≤ (k + 16)/1.21] ≈ 0.1546 + 0.786

Pr[Z ≤ (k + 16)/1.21] ≈ 0.940

Take the inverse CDF of both sides (Φ(x) denotes the CDF itself):

(k + 16)/1.21 ≈ Φ⁻¹ (0.940) ≈ 1.556

Solve for k :

k + 16 = 1.21 • 1.556

k ≈ -14.118

8 0

3 years ago

Coach Jeffries had forty-six students sign up for after-school soccer. Then some students decided to join after-school basketbal

Natasha_Volkova [10]

21

hwhdidjcbudhdhdhdvdhkd

7 0

3 years ago

Electronics Unlimited sells TVs. There are 110 TVs on display in the showroom, and each TV is turned on to a random channel from

Lynna [10]

Answer:

Option B. There will be close to 40 TVs but probably not exactly 40 TVs not showing a sports channel.

Explanation:

There are a total of 7 sports channels and 4 non-sports channels out of 11 channels.

The probability that one TV will be not be showing a sports channel, P(not S), is:

P(not S) = number of non-sports channels / number of channels.

P(not S) = 4/11.

The best prediction on how many TVs will not be showing a sports channel is, the expected value, which is equal to the number of TVs mulitplied by P(not S):

P(not S) = 110 × 4/11 = 40.

Since this is a random variable, the expected value is not the exact number of TV but just a probability.

Hence, the answer is the option B: There will be close to 40 TVs but probably not exactly 40 TVs not showing a sports channel.

3 0

3 years ago

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