Can someone please help me with these 3 problems..

Kendall and Asa are both saving up money to buy new mountain bikes. Kendall has $50 and will save $10 every week. Asa has $15 an

zloy xaker [14]

Answer: 7 weeks

Step-by-step explanation:

6 0

3 years ago

1 in on the map equals 16 miles, how many in on the map would represent 24 miles ?

Paul [167]

1.5 or 1 1/2 in. On the map equals24 miles

6 0

4 years ago

What unit would you measure a pitcher of iced tea in?

wel

Pounds.. or grams it depends on what size the pitcher is.. but pounds

5 0

4 years ago

Read 2 more answers

A computer chess game and a human chess champion are evenly matched. They play ten games. Find probabilities for the following e

AleksAgata [21]

Answer:

(a) The probability that each of them win five games is 0.2461.

(b) The probability that the computer wins seven games is 0.1172.

(c) The probability that the human wins at least 7 games is 0.1711.

Step-by-step explanation:

Let the random variable X = the human wins a chess game.

The probability that the human wins a game is, P (X) = p = 0.50.

The number of games played by the computer and the human is, n = 10.

The random variable $X\sim Bin(n = 10,p=0.50)$

The probability distribution of the Binomial random variable X is:

$P (X=x)={n\choose x}p^{x}(1-p)^{n-x}={10\choose x}(0.50)^{x}(1-0.50)^{10-x}$

(a)

If both the computer and the human wins five games each then the probability of the human winning 5 games is:

$P (X=5)={10\choose 5}(0.50)^{5}(1-0.50)^{10-5}\\=252\times 0.03125\times0.03125\\=0.246094\\\approx 0.2461$

Thus, the probability that each of them win five games is 0.2461.

(b)

If the computer wins 7 games then the number of games won by the human is, 10 - 7 = 3.

P (Computer winning 7 games) = P (Human winning 3 games)

The probability that the human wins 3 games is:

$P (X=3)={10\choose 3}(0.50)^{3}(1-0.50)^{10-3}\\=120\times 0.125\times0.0078125\\=0.1171875\\\approx 0.1172$

Thus, the probability that the computer wins seven games is 0.1172.

(c)

Compute the probability that the human wins at least 7 games as follows:

$P(X\geq 7)=P(X=7)+P(X=8)+P(X=9)+P(X=10)\\={10\choose 7}(0.50)^{7}(1-0.50)^{10-7}+{10\choose 8}(0.50)^{8}(1-0.50)^{10-8}\\+{10\choose 9}(0.50)^{9}(1-0.50)^{10-9}+{10\choose 10}(0.50)^{10}(1-0.50)^{10-10}\\=0.1172+0.044+0.009+0.0009\\=0.1711$

Thus, the probability that the human wins at least 7 games is 0.1711.

5 0

4 years ago

I just need to see how to work this problem out, so I don't have to ask a question like this again..and I know what I'm doing ne

mihalych1998 [28]

Answer by Mimiwhatsup:

$-6x+12>24\\\mathrm{Subtract\:}12\mathrm{\:from\:both\:sides}\\-6x+12-12>24-12\\Simplify\\-6x>12\\Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)\\\left(-6x\right)\left(-1\right)$

$=-\frac{12}{6}\\\mathrm{Divide\:the\:numbers:}\:\frac{12}{6}=2\\=-2\\Answer is: x$

8 0

3 years ago