If a sector of a circle has a radius of 15m and an arc length of 25 meters, find the central

A family plans to have 3 children. For each birth, assume that the probability of having a boy is the same as the probability of

Rzqust [24]

Answer:

1/2

Step-by-step explanation:

50-50 chance they will have a boy each time, 50-50 chance they will have a girl each time

3 0

3 years ago

There was 2/3 of a pan of a lasagna in the refrigerator. Bill and his friends ate half of what was left. Write a number sentence

olga nikolaevna [1]

Bill and his friends ate 1/3 of the pan, you get this by doing: 2/3 / 1/2 = 1/3 or 0.3333

7 0

2 years ago

He difference of 29 and n

almond37 [142]

Answer:

28

Step-by-step explanation:

n can be represented as 1.

29 - 1 = 28

Please visit my church's website at fbcinterlaken.org to learn more about God. Please subscribe to the You Tube channel because my church really needs the support right now. Thank you! And have a nice day! :) (I will take brainliest).

3 0

2 years ago

Read 2 more answers

Help! Will give brainliest and 5 stars rating and thanks

Morgarella [4.7K]

Answer:

x < -4 or x > 8

On a number line, make two arrows:

1) going towards left from -4 with an open circle at -4

2) going towards right from 8 with an open circle at 8

Step-by-step explanation:

l -4 +2x l -3 > 9

l -4 +2x l > 12

l 2x - 4 l > 12

2x - 4 > 12

2x > 16

x > 8

-2x + 4 > 12

-2x > 8

x < -4

On a number line, make two arrows:

1) going towards left from -4 with an open circle at -4

2) going towards right from 8 with an open circle at 8

8 0

3 years ago

Read 2 more answers

The U.S. government has devoted considerable funding to missile defense research over the past 20 years. The latest development

Bad White [126]

Answer:

a) Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

b) $X \sim Bin(n =20, p=0.8)$

c) $P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

d) $P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

e) $E(X) = np = 20*0.8 = 16$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

Part b

$X \sim Bin(n =20, p=0.8)$

Part c

For this case we just need to replace into the mass function and we got:

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

Part d

For this case we want this probability: $P(X\geq 15)$

And we can solve this using the complement rule:

$P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

Part e

The expected value is given by:

$E(X) = np = 20*0.8 = 16$

5 0

3 years ago