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Cerrena [4.2K]
1 year ago
9

On a number line, point F is at 4, and point G is at -2. Point H lies between point F and point G. If the ratio of FH to HG is 3

:9, where does point H lie on the number line?
Mathematics
1 answer:
Mademuasel [1]1 year ago
7 0

Here, we are required to determine the point where point H lie on the number line.

The point H is at point 2.5 on the number line.

The distance between point F and G on the number line is:

4 - (-2) = 6 numbers on the number line.

And since the ratio of FH to HG is 3:9

  • Then, let point H be at no. X on the number line.

Therefore; FH/HG = (4 - x)/(x - (-2))

i.e 3/9 = (4 - x)/ (x +2).

By cross multiplication, 3x + 6 = 36 - 9x.

Therefore, 12x = 30

and, x = 2.5

Therefore, the point H is at point 2.5 on the number line.

Read more:

brainly.com/question/18329661

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Step-by-step explanation:

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The aquarium at Sea Critters Depot contains 140 fish. Eighty of these fish are green swordtails (44 female and 36 male) and 60 a
maxonik [38]

Given that:

Total number of fish = 140

Fish are green swordtails female = 44

Fish are green swordtails male = 36

Fish are orange swordtails female = 36

Fish are orange swordtails male = 24

Solution:

A. We have to find the probability that the selected fish is a green swordtail.

\text{P(green swordtail)}=\dfrac{\text{Total green swordtail fish}}{\text{Total fish}}

\text{P(green swordtail)}=\dfrac{80}{140}

\text{P(green swordtail)}=\dfrac{4}{7}

Therefore, the probability that the selected fish is a green swordtail is \dfrac{4}{7}.

B.  We have to find the probability that the selected fish is male.

\text{P(Male fish)}=\dfrac{\text{Total male fish}}{\text{Total fish}}

\text{P(Male fish)}=\dfrac{36+24}{140}

\text{P(Male fish)}=\dfrac{60}{140}

\text{P(Male fish)}=\dfrac{3}{7}

Therefore, the probability that the selected fish is a male, is \dfrac{3}{7}.

C. We have to find the probability that the selected fish is a male green swordtail.

\text{P(Male green swordtail)}=\dfrac{\text{Total male green swordtail fish}}{\text{Total fish}}

\text{P(Male green swordtail)}=\dfrac{36}{140}

\text{P(Male green swordtail)}=\dfrac{9}{35}

Therefore, probability that the selected fish is a male green swordtail is \dfrac{9}{35}.

D.

We have to find the probability that the selected fish is either a male or a green swordtail.

\text{P(Male or green swordtail)}=\dfrac{\text{Total male or green swordtail fish}}{\text{Total fish}}

\text{P(Male or green swordtail)}=\dfrac{44+36+24}{140}

\text{P(Male or green swordtail)}=\dfrac{96}{140}

\text{P(Male or green swordtail)}=\dfrac{24}{35}

Therefore, the probability the selected fish is either a male or a green swordtail is \dfrac{24}{35}.

4 0
2 years ago
5. In a big, red box, there are 7 smaller blue boxes. In each of
Alex787 [66]

Answer:

2801 boxes

Step-by-step explanation:

You have 1 red box

You have 7 blue boxes, so far 8 boxes

There are 49 total black boxes (7*7), making 57 total boxes

There are 49*7 = 343 yellow boxes, making 400 boxes

There are 343*7 = 2401 gold boxes, making for a total of 2801 boxes.

7 0
1 year ago
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