Ask question

Ask question

All categories

3 years ago

14

Jamal bikes 5 miles in 20 minutes, or of an

1 answer:

anyanavicka [17]3 years ago

5 0

Jamal would bike further in

You might be interested in

Find the Value of Y! Thank u!!!

Virty [35]

Answer:

my bad if im wrong but i got y=35

Step-by-step explanation:

4 0

3 years ago

A coin is tossed 10 times resulting in 7 heads and 3 tails. The same coin is tossed 1000 times resulting in 510 heads and 490 ta

KiRa [710]

The answer is <span>C. 50%.

The theoretical probability has nothing to do with the experiments. So, we will forget results of the experiment and think about theoretical probability. A coin has two sides - head and tail. The probability to get head is 1/2 = 0.5 = 50%. This is because if you toss the coin and you get head, head is one probability of two probability in total (head and tail). The same situation is with tail. Tail is .</span><span>one probability of two probability in total (head and tail).</span>

8 0

3 years ago

In Game 1, Emerson struck out 30 times in 90 times at bat. In Game 2, he struck out 40 times in 120 times at bat. In Game 3, Eme

VikaD [51]

Answer:

<u>Total number of strikeouts:</u>

30 + 40 + 42 = 112

<u>Total number of at bats:</u>

90 + 120 + 140 = 350

<u>Hits = times - strikeouts:</u>

350 - 112 = 238

<u>Percentage:</u>

238/352*100% = 67.61%

7 0

3 years ago

Me trying to pass test

user100 [1]

Answer:

b

Step-by-step explanation:

b is the answer to your test

6 0

3 years ago

Read 2 more answers

For about $1 billion in new space shuttle expenditures, NASA has proposed to install new heat pumps, power heads, heat exchanger

11111nata11111 [884]

Answer:

The probability of one or more catastrophes in:

(1) Two mission is 0.0166.

(2) Five mission is 0.0410.

(3) Ten mission is 0.0803.

(4) Fifty mission is 0.3419.

Step-by-step explanation:

Let <em>X</em> = number of catastrophes in the missions.

The probability of a catastrophe in a mission is, P (X) = $p=\frac{1}{120}$ .

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:

$P(X=x)={n\choose x}\frac{1}{120}^{x}(1-\frac{1}{120})^{n-x};\x=0,1,2,3...$

In this case we need to compute the probability of 1 or more than 1 catastrophes in <em>n</em> missions.

Then the value of P (X ≥ 1) is:

P (X ≥ 1) = 1 - P (X < 1)

= 1 - P (X = 0)

$=1-{n\choose 0}\frac{1}{120}^{0}(1-\frac{1}{120})^{n-0}\\=1-(1\times1\times(1-\frac{1}{120})^{n-0})\\=1-(1-\frac{1}{120})^{n-0}$

(1)

Compute the compute the probability of 1 or more than 1 catastrophes in 2 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{2-0}=1-0.9834=0.0166$

(2)

Compute the compute the probability of 1 or more than 1 catastrophes in 5 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{5-0}=1-0.9590=0.0410$

(3)

Compute the compute the probability of 1 or more than 1 catastrophes in 10 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{10-0}=1-0.9197=0.0803$

(4)

Compute the compute the probability of 1 or more than 1 catastrophes in 50 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{50-0}=1-0.6581=0.3419$

6 0

3 years ago