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3 years ago

Can any one do B and B please help me I need help me please thank you

Mathematics

1 answer:

MissTica3 years ago

3 0

Answer:

see below

Step-by-step explanation:

1 B Answer :

$7\times7\times7\times7\times7/7\times7\times7\times7=7$ or $7^{2+3}/7^{4} =7^{5-4} =7^{1} =7$

2 B Answer :

$k^{4\times2} \times k^{5} =k^{8+5} =k^{13}$

3B Answer :

$5a-15-a/1=5a-a-15=4a-15$

Hope that help

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the sum of two numbers is 3 3/4 if 1/3 of the larger number is 1/2 more than 2/3 of the smaller find the number

Agata [3.3K]

Let us assume the larger number = x
Let us assume the smaller number = y
Then
x + y = 3 3/4
x + y = 15/4
And
x/3 = (2y/3) + 1/2
x = [3 * (2y/3)] + (3/2)
= 2y + (3/2)
Now putting the value of x from the second equation to the first , we get
x + y = 15/4
2y + (3/2) + y = 15/4
3y = (15/4) - (3/2)
3y = (15 - 6)/4
3y * 4 = 9
12y = 9
y = 9/12
   = 3/4
Now putting the value of y in the first equation, we get
x + y = 15/4
x + (3/4) = (15/4)
x = (15/4) - (3/4)
   = (15 - 3)/4
   = 12/4
   = 3
So the value of x or the larger number is 3 and the value of y or the smaller number is 3/4.

3 0

2 years ago

According to government data, 20% of employed women have never been married. If 10 employed women are selected at random, what i

Ierofanga [76]

Answer:

a) $P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

b) $P(X\leq 2) = P(X=0) + P(X=1) +P(X=2)$

$P(X=0) = (10C0) (0.2)^0 (1-0.2)^{10-0}= 0.107$

$P(X=1) = (10C1) (0.2)^1 (1-0.2)^{10-1}= 0.268$

$P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

And replacing we got:

$P(X\leq 2) = 0.107+0.268+0.302=0.678$

c) For this case we want this probability:

$P(X\geq 8) = P(X=8) + P(X=9) +P(X=10)$

But for this case the probability of success is p =1-0.2= 0.8

We can find the individual probabilities and we got:

$P(X=8) = (10C8) (0.8)^8 (1-0.8)^{10-8} =0.302$

$P(X=9) = (10C9) (0.8)^9 (1-0.8)^{10-9} =0.268$

$P(X=10) = (10C10) (0.8)^{10} (1-0.8)^{10-10} =0.107$

And replacing we got:

$P(X \geq 8) = 0.677$

And replacing we got:

$P(X\geq 8)=0.0000779$

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".

Let X the random variable of interest, on this case we now that:

$X \sim Bin (n=10 ,p=0.2)$

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:

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

Solution to the problem

Let X the random variable "number of women that have never been married" , on this case we now that the distribution of the random variable is:

$X \sim Binom(n=10, p=0.2)$

Part a

We want to find this probability:

$P(X=2)$

And using the probability mass function we got:

$P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

Part b

For this case we want this probability:

$P(X\leq 2) = P(X=0) + P(X=1) +P(X=2)$

We can find the individual probabilities and we got:

$P(X=0) = (10C0) (0.2)^0 (1-0.2)^{10-0}= 0.107$

$P(X=1) = (10C1) (0.2)^1 (1-0.2)^{10-1}= 0.268$

$P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

And replacing we got:

$P(X\leq 2) = 0.107+0.268+0.302=0.678$

Part c

For this case we want this probability:

$P(X\geq 8) = P(X=8) + P(X=9) +P(X=10)$

But for this case the probability of success is p =1-0.2= 0.8

We can find the individual probabilities and we got:

$P(X=8) = (10C8) (0.8)^8 (1-0.8)^{10-8} =0.302$

$P(X=9) = (10C9) (0.8)^9 (1-0.8)^{10-9} =0.268$

$P(X=10) = (10C10) (0.8)^{10} (1-0.8)^{10-10} =0.107$

And replacing we got:

$P(X \geq 8) = 0.677$

3 0

3 years ago

Select the correct answer.

zepelin [54]

Answer:

Answer is option c

Step-by-step explanation:

$given \: equations \: are \\ x - 2y = 15 \: \: and \: 2x + 4y = - 18 \\ divide \: by \: 2 \: in \: second \: equation \\ x + 2y = - 9 \: \: (equation \: 3) \\ adding \: 1and \: 3 \: we \: get \\ 2x = 6 \\ x = 3 \\ substitute \: x = 3 \: in \: equation \: 3 \\ 3 + 2y = - 9 \\ 2y = - 9 - 3 \\ 2y = - 12 \\ y = - 6$