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grigory [225]
3 years ago
6

Is 5 a factor of 35 56 51 40

Mathematics
2 answers:
tester [92]3 years ago
6 0
5 is factor of 35 and 40 not 56 and 51
FromTheMoon [43]3 years ago
3 0
5 is a factor of 35 and 40 but not 56 and 51. Factors of 5 always end in either 0 or 5.
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Ria can paint a room in 4 hours. Destiny can paint the same room in 6 hours. How long would it take Ria and Destiny to paint the
Lerok [7]

Time taken by Ria to paint a room = 4 hours

Time taken by Destiny to paint a room = 6 hours

If they work together, they complete 1 job. So,

\frac{1x}{4}+\frac{1x}{6}=1

The LCD of 4 and 6 is 12

\frac{3x+2x}{12}=1

\frac{5x}{12}=1

5x=12

x=\frac{12}{5}

Hence, both of them can paint the room in \frac{12}{5} hours or 2.4 hours.


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How many zeros does this polynomial function, y=(x-8)(x+3)^2
lidiya [134]

Answer:

3

Step-by-step explanation:

The factor (x - 8) has 1 zero at x = 8

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In total there are 3 zeros


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3 years ago
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Finger [1]
The ratio is 12:28. Or in fraction form it’s 12/28. You can also simplify it to 3/7
4 0
3 years ago
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An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.
andrey2020 [161]

Answer:

The probability that A selects the first red ball is 0.5833.

Step-by-step explanation:

Given : An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.

To find : What is the probability that A selects the first red ball?

Solution :

A wins if the first red ball is drawn 1st,3rd,5th or 7th.

A red ball drawn first, there are E(1)= ^9C_2 places in which the other 2 red balls can be placed.

A red ball drawn third, there are E(3)= ^7C_2 places in which the other 2 red balls can be placed.

A red ball drawn fifth, there are E(5)= ^5C_2 places in which the other 2 red balls can be placed.

A red ball drawn seventh, there are E(7)= ^3C_2 places in which the other 2 red balls can be placed.

The total number of total event is S= ^{10}C_3

The probability that A selects the first red ball is

P(A \text{wins})=\frac{(^9C_2)+(^7C_2)+(^5C_2)+(^3C_2)}{^{10}C_3}

P(A \text{wins})=\frac{36+21+10+3}{120}

P(A \text{wins})=\frac{70}{120}

P(A \text{wins})=0.5833

6 0
3 years ago
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