All categories

1 year ago

Pat's income is 20 % more than Adam. How much percent is Adam's income less than Pat's?

Mathematics

1 answer:

pantera1 [17]1 year ago

4 0

Answer: 20%

Step-by-step explanation: The same number applies here. Think about it this way, let's say you have a friend named Sofia and she has 3 dollars more than you, how many dollars do you have less than Sofia? 3 dollars. That's the easiest way I can explain it.

You might be interested in

A baker records the internal temperature of a pie that has been left out on the counter. The room temperature is 14 Celsius. An

maksim [4K]

Assume r= t not sure if you mistyped this.
t = 15 min

T(t) = 68 (0.5)^(15/10) = 24 Celsius

31 = (0.5)^(t/10) take ln

ln31 = (t/10) ln (0.5) + ln68

t = (ln32 - ln 68/ln0.5) * 10 = 11.3 min

5 0

2 years ago

A gas can holds 10liters of gas. How many cans could we fill with 7 liters of gas?

telo118 [61]

Answer:

yes agas can hold 10 l of gas

Step-by-step explanation:

.comvmessage on it

6 0

1 year ago

A large manufacturing plant has analyzed the amount of time required to produce an electrical part and determined that the times

WARRIOR [948]

Answer:

We conclude that the new procedure will not decrease the population mean amount of time required to produce the part.

Step-by-step explanation:

We are given that a large manufacturing plant has analyzed the amount of time required to produce an electrical part and determined that the times follow a normal distribution with mean time μ = 45 hours.

A random sample of 25 parts will be selected and the average amount of time required to produce them will be determined. The sample mean amount of time is = 43.118 hours with the sample standard deviation s = 5.5 hours

Let $\mu$ = population mean amount of time required to produce an electrical part using new procedure

SO, Null Hypothesis, $H_0$ : $\mu \geq$ 45 hours {means that the new procedure will remain same or increase the population mean amount of time required to produce the part}

Alternate Hypothesis, $H_a$ : $\mu$ < 45 hours {means that the new procedure will decrease the population mean amount of time required to produce the part}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }$ ~ $t_n_-_1$

where, $\mu$ = sample mean amount of time = 43.118 hours

s = sample standard deviation = 5.5 hours

n = sample of parts = 25

So, test statistics = $\frac{43.118-45}{{\frac{5.5}{\sqrt{25} } } }$ ~ $t_2_4$

= -1.711

Now at 0.025 significance level, the t table gives critical value of -2.064 at 24 degree of freedom for left-tailed test. Since our test statistics is more than the critical value of t so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the new procedure will remain same or increase the population mean amount of time required to produce the part.

4 0

1 year ago

HELP ASAP!!! (25 points and I’ll mark brainliest!!!!)

defon

Answer:

1. Take the Average of the distances the ball travelled each hit.

2. He should use the Interquartile Range. This is the difference between the Upper Quartile and the Lower Quartile of the distances he hits the ball.

3. He should use Mean

4. He should use Median. It best measures skewed data

Step-by-step explanation:

THE FIRST PART.

Raul should take the average of the distances the ball travelled each hit.

This is done by summing the total distances the ball travelled each bounce, and then dividing the resulting value by the total number of times he hit the ball, which is 10.

THE SECOND PART

He should use the Interquartile Range. This is the difference between the Upper Quartile and the Lower Quartile of the distances he hits the ball.

THE THIRD PART

He should take the mean of the distances of the ball that stayed infield.

This is the distance that occurred the most during the 9 bounces that stayed infield. The one that went outfield is makes it unfair to use any other measure of the center, taking the mean will give a value that is significantly below his efforts.

THE FOURTH PART

He should take the Median of the data, it is best for skewed data.

This is the middle value for all the distances he recorded.

3 0

1 year ago

What do you call it when 50 people stand on a wooden dock

umka2103 [35]

here's a worksheet that might match what you're looking for

6 0

1 year ago