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amm1812
3 years ago
5

This one to help ASAP !

Mathematics
2 answers:
pantera1 [17]3 years ago
7 0

Answer:

C

Step-by-step explanation:

Hoochie [10]3 years ago
4 0
I’m pretty sure it’s “C”
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Which of the following shows the number of time 3 1/5 fits into 10 1/2
ollegr [7]
The answer is 0.304762
7 0
4 years ago
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Will give brainliest
xeze [42]

Answer:

<h2><u>7</u>x - <u>7</u>y + <u>21</u> </h2>

Step-by-step explanation:

7(x - y + 3)

=> 7 × x - 7 × y + 7 × 3

=> <u>7</u>x - <u>7</u>y + <u>21</u>

6 0
3 years ago
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Yahoo creates a test to classify emails as spam or not spam based on the contained words. This test accurately identifies spam (
Amiraneli [1.4K]

Answer and Step-by-step explanation:

The computation is shown below:

Let us assume that

Spam Email be S

And, test spam positive be T

Given that

P(S) = 0.3

P(\frac{T}{S}) = 0.95

P(\frac{T}{S^c}) = 0.05

Now based on the above information, the probabilities are as follows

i. P(Spam Email) is

= P(S)

= 0.3

P(S^c) =  1 - P(S)

= 1 - 0.3

= 0.7

ii. P(\frac{S}{T}) = \frac{P(S\cap\ T}{P(T)}

= \frac{P(\frac{T}{S}) . P(S) }{P(\frac{T}{S}) . P(S) + P(\frac{T}{S^c}) . P(S^c) }

= \frac{0.95 \times 0.3}{0.95 \times 0.3 + 0.05 \times 0.7}

= 0.8906

iii. P(\frac{S}{T^c}) = \frac{P(S\cap\ T^c}{P(T^c)}

= \frac{P(\frac{T^c}{S}) . P(S) }{P(\frac{T^c}{S}) . P(S) + P(\frac{T^c}{S^c}) . P(S^c) }

= \frac{(1 - 0.95)\times 0.3}{ (1 -0.95)0.95 \times 0.3 + (1 - 0.05) \times 0.7}

= 0.0221

We simply applied the above formulas so that the each part could come

8 0
3 years ago
For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4
Elina [12.6K]

Answer:

<em>C.</em> (5-\frac{1}{2})^6

Step-by-step explanation:

Given

15(5)^2(-\frac{1}{2})^4

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

Sum = 2 + 4

Sum = 6

Each term of a binomial expansion are always of the form:

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

Where n = the sum above

n = 6

Compare 15(5)^2(-\frac{1}{2})^4 to the above general form of binomial expansion

(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......

Substitute 6 for n

(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......

[Next is to solve for a and b]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

By direct comparison of

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

and

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

We have;

^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4

Further comparison gives

^nC_r = 15

a^{n-r} =(5)^{6-4}

b^r= (-\frac{1}{2})^4

[Solving for a]

By direct comparison of a^{n-r} =(5)^{6-4}

a = 5

n = 6

r = 4

[Solving for b]

By direct comparison of b^r= (-\frac{1}{2})^4

r = 4

b = \frac{-1}{2}

Substitute values for a, b, n and r in

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

Solve for ^6C_4

(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em>(5-\frac{1}{2})^6<em />

3 0
3 years ago
Create an equivalent system of these equations and test your solution <br> x + y = 1<br> x 3y =9
ad-work [718]
An equvilent equation
remember you can do anything to an equation as long asyou do it to both sides


assuming yo have
x+y=1 and
x-3y=9
mulitply both by 2
2x+2y=2
2x-6y=18
those are equvilent



ok, solve initial

x+y=1
x-3y=9
multiply first equation by -1 and add to 2nd equation


-x-y=-1
<u>x-3y=9 +</u>
0x-4y=8

-4y=8
divide both sides by -4
y=-2

sub back
x+y=1
x-2=1
add 2
x=3


x=3
y=-2
(3,-2)

if we test it in other one

2x+2y=2
2(3)+2(-2)=2
6-4=2
2=2
yep

2x-6y=18
2(3)-6(-2)=18
6+12=18
18=18
yep


solution is (3,-2)
4 0
3 years ago
Read 2 more answers
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