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Greeley [361]
3 years ago
15

Jerry is twice as old as Marty and Phil is 5 years old than Marty. Altogether their ages add

Mathematics
1 answer:
KATRIN_1 [288]3 years ago
7 0

Answer:

Jerry is 90, Marty 45 and Phil is 50

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−3/8 ÷ 1/2= the numbers are fractions
marishachu [46]

\text {Hi! To solve this problem you must...}

\text {Do Keep Change Flip (KCF):}

\text {Keep: -3/8}

\text {Change: /  into *}

\text {Flip: 1/2 into 2/1}

\text {Your new problem should be -3/8*2/1}

\text {The Final Step is to Multiply:}

\text {Note: When Multiplying Fractions you multiply Numerator by Numerator}

\text {and Denominator by Denominator}

\text {-3/8*2/1=}

\fbox {-6/8}

\text {Best of Luck!}

\text {-LimitedX}

4 0
3 years ago
A box contains different colored paper clips. The probability of drawing two red paper clips from the box without replacement is
ivann1987 [24]

Answer: 5/14 which is choice B

================================================

How I got this answer:

Define the following events

A = event of picking a red paper clip on the first selection

B = event of picking a red paper clip on the second drawing

Replacement is not made.

Now onto the probabilities for each

P(A) = 2/5 = 0.4 is given to us as this is simply the probability of picking red on the first try

P(A and B) = probability of both events A and B happeing simultaneously = 1/7

P(B|A) = probability event B occurs, given event A has occured

P(B|A) = probability of selecting red on second selection, given first selection is red (no replacement)

P(B|A) = P(A and B)/P(A)

P(B|A) = (1/7) / (2/5)

P(B|A) = (1/7) * (5/2)

P(B|A) = (1*5)/(7*2)

P(B|A) = 5/14

So if event A happens, then the chances of event B happening is 5/14

------------------

A more concrete example:

If we had 15 paperclips, and 6 of them were red, then

P(A) = (# of red)/(# total) = 6/15 = 2/5

P(B|A) = (# of red left)/(# total left) = (6-1)/(15-1) = 5/14

P(A and B) = P(A)*P(B|A) = (2/5)*(5/14) = 10/70 = 1/7

7 0
4 years ago
Read 2 more answers
2x^3-x^2+x-2 factorize​
bekas [8.4K]

Answer:

(x-1)(2x^2+x+2)

Step-by-step explanation:

Factorize:

f(x)=2x^3-x^2+x-2

<u>Factor Theorem</u>

If f(a) = 0 for a polynomial then (x - a) is a factor of the polynomial f(x).

Substitute x = 1 into the function:

\implies f(1)=2(1)^3-1^2+1-2=0

Therefore, (x - 1) is a factor.

As the polynomial is cubic:

\implies  f(x)=(x-1)(ax^2+bx+c)

Expanding the brackets:

\implies  f(x)=ax^3+bx^2+cx-ax^2-bx-c

\implies  f(x)=ax^3+(b-a)x^2+(c-b)x-c

Comparing coefficients with the original polynomial:

\implies ax^3=2x^3 \implies a=2

\implies (b-a)x^2=-x^2 \implies b-2=-1 \implies b=1

\implies -c=-2 \implies c=2

Therefore:

\implies  f(x)=(x-1)(2x^2+x+2)

Cannot be factored any further.

4 0
2 years ago
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Plssss help (ALGEBRA 1 ) will mark brainlest
Dimas [21]

Answer:

I'm confused by the layout of the first two, but I can help you with the last one!

The last one is C!

Step-by-step explanation:

7 0
3 years ago
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The sum of two numbers is 27. One number is 3 more than the other. Write and solve a system of equations to find the two numbers
givi [52]
X + y = 27
x = y + 3

y + 3 + y = 27
2y + 3 = 27
2y = 27 - 3
2y = 24
y = 24/2
y = 12

x = y + 3
x = 12 + 3
x = 15

ur numbers are 12 and 15
7 0
4 years ago
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