All categories

3 years ago

Choose the set of equivalent fractions that correctly uses the LCD for these fractions 3/4 and 1/6

Mathematics

1 answer:

valina [46]3 years ago

6 0

Answer: 9/12, 2/12

Step-by-step explanation:

Start by writing multiples of 6 and 4

4: 4, 8, 12

6: 6, 12, 18

Did you see the multiple that both numbers have? Yup, it's 12. So that has to be your denominator for both fractions. This is because you have to have the LEAST COMMON Denominator (LCD). So both numbers have 12 in common as one of their multiples and, it's the least common one because there are no numbers that come before it that are common for both numbers. Now, you just have to change the denominators of both fractions into 12. To do that to 3/4, just multiply the top and bottom by 3 to get 9/12. To do that to 1/6, just multiply the top and bottom by 2 to get 2/12. Hope this helps!

NOTE: REMEBER THAT WHATEVER YOU DO TO THE TOP OF A FRACTION, YOU HAVE TO DO IT TO THE BOTTOM TOO (AND VICE VERSA)

You might be interested in

Pls help no links in the answer

Dominik [7]

Answer:

Hello my friend, I hope this helps you!

Step-by-step explanation:

7 0

3 years ago

Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis

lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = the interval of time between the eruption

So, X ~ Normal( $\mu=75, \sigma^{2} =20$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

P(X > 84 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{84-75}{20}$ ) = P(Z > 0.45) = 1 - P(Z $\leq$ 0.45)

= 1 - 0.6736 = 0.3264

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{13} } }$ ) = P(Z > 1.62) = 1 - P(Z $\leq$ 1.62)

= 1 - 0.9474 = 0.0526

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{20} } }$ ) = P(Z > 2.01) = 1 - P(Z $\leq$ 2.01)

= 1 - 0.9778 = 0.0222

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0

3 years ago

19.2 is what percent of 24?

Pie

Answer: 9.2 is 80% of 24.

hope this helps!

7 0

3 years ago

BabaBlast [244]

Answer: 14.88 is the answer

Step-by-step explanation:

8 0

3 years ago

never [62]

Answer:

○ B. $525\:square\:units$

Step-by-step explanation:

All sides meet at right angles, so when split up, you have this:

$525 = 98 + 21 + 154 + 252 = 7 \times 14 + 3 \times 7 + 11 \times 14 + 9 \times 28$

I am joyous to assist you anytime.

8 0

3 years ago