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3 years ago

Can someone please so me how I would figure out how to do this? I believe it’s subtraction. But I’m not sure

Mathematics

1 answer:

elena-s [515]3 years ago

6 0

Answer:

A. 1/12.

Step-by-step explanation:

Orville and Wilbur ate 1/3 + 1/4

= 4/12 + 3/12

= 7/12 of the pizza.

So there was 1 - 7/12 = 5/12 of the pizza left.

So Ruth ate 5/12 * 1/5

= 5/60

= 1/12.

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i need help answering this but also how to do it like step by step. i don’t understand this. (9 + 7x) - 7x

anyanavicka [17]

Answer: 9

Step-by-step explanation: so first you remove the ( ) because they’re useless in this then once you remove then Since two opposites add up to zero, remove them from the expression then you only get left with 9.

7 0

3 years ago

In which quadrant is the number 6 – 8i located on the complex plane?

antoniya [11.8K]

The number 6 – 8i is located in the 4th quadrant on the complex plane.

4 0

3 years ago

Read 2 more answers

The Dowers make pumpkin pie. If each person wants three slices of pie, and each pie is Cut into 9 slices, how many pies do they

Mrac [35]

The Dowers need to make 5 pies for 15 people.

Step-by-step explanation:

Given,

1 person = 3 slices of pie

15 persons = 15*3 = 45 slices of pie

It is also given that;

9 slices of pie = 1 whole pie

1 slice of pie = $\frac{1}{9}\ pie$

45 slices of pie = $\frac{1}{9}*45$

45 slices of pie = 5 whole pies

The Dowers need to make 5 pies for 15 people.

Keywords: multiplication, division

Learn more about multiplication at:

#LearnwithBrainly

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

Find the face value of the 20-year zero-coupon bond at 4.4%, compounded semiannually, with a price of $8,375.

umka21 [38]

Given:

Time,

t = 20 years

Rate,

r = 4.4%

Price

= $8,375

Now,