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blsea [12.9K]
2 years ago
15

A monsoon dumped rain on a coastal area. in twelve hours 20 inches of rain had fallen. how much rain will fall over a period of

2 days, if it continues to rain at a constant rate?​
Mathematics
2 answers:
marissa [1.9K]2 years ago
6 0

Answer:

80

Step-by-step explanation:

I asked my brother and he he told me

Setler [38]2 years ago
4 0

Answer:

80

Step-by-step explanation:

If in 12 hours 20 inches falls in 24 hours 40 inches will fall and another 12 hours would be 60 inches. And another 12 hours you will get 2 days so your answer is 80

You might be interested in
I was told that if I had perfect attendance, I would get an extra 4% in my check as a bonus. Therefore, if I earned $960.00 befo
MakcuM [25]

Answer:

Your total income with perfect attendance would be $998.40

Step-by-step explanation:

First you need to figure out how much the 4% bonus is by multiplying 4% by $960.

But first you need to convert 4% to a decimal so it becomes 0.04

Then multiply it by $960 which would get you $38.40

So the 4% bonus is $38.40 but you need to know how much your total income is with the bonus.

So you need to add $38.40 and $960 which will get you $998.40

The total income with perfect attendance would be $998.40

3 0
2 years ago
It is believed that as many as 23% of adults over 50 never graduated from high school. We wish to see if this percentage is the
JulijaS [17]

Answer:

1)  n=48  

2) n=298

3) n=426

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

p represent the real population proportion of interest

\hat p represent the estimated proportion for the sample

n is the sample size required (variable of interest)

z represent the critical value for the margin of error

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

Part 1

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by \alpha=1-0.90=0.10 and \alpha/2 =0.05. And the critical value would be given by:  

z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64  

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}} (a)  

And on this case we have that ME =\pm 0.1 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2} (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:  

n=\frac{0.23(1-0.23)}{(\frac{0.1}{1.64})^2}=47.63  

And rounded up we have that n=48  

Part 2

The margin of error on this case changes to 0.04 so if we use the same formula but changing the value for ME we got:

n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.64})^2}=297.7  

And rounded up we have that n=298  

Part 3

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:  

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96  

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}} (a)  

And on this case we have that ME =\pm 0.04 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2} (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:  

n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.96})^2}=425.22  

And rounded up we have that n=426  

3 0
2 years ago
A process engineer is implementing a quality assurance system on a breakfast cereal production line. A new sensor is installed o
nataly862011 [7]

Answer:

0.2%

Step-by-step explanation:

The given parameters are;

The percentage of product within the correct range = 95%

The percentage of the times the sensor reject boxes of incorrect weight = 98%

The percentage of the times the company reject boxes with correct weight = 1%

\begin{array}{cccc}&P&N&T\\P&TP&0.9&90\\N&FP&9.8&10\end{array}

We have, FN = 0.9, TN = 9.8, TP = 90 - 0.9 = 89.1, FP = 10 - 9.8 = 0.2

FNR = FN/(FN + TP) = 0.9/(0.9 + 89.1) = 0.01

FPR = 0.2/(0.2 + 9.8) = 0.02

TPR = 89.1/(89.1 + 0.9) = 0.99

TNR = 9.8/(9.8 + 0.2) = 0.98

P(C/A) = TPR × P(C)/((TPR × P(C) + FPR×(1 - P(C)))

Where;

P(C/A) = The probability that a correct weight package is accepted

∴ P(C/A) = 0.99*0.9/(0.99*0.9 + 0.02*0.1) ≈ 0.99776

The probability that a correctly weighed package is rejected, P(C/R), is given as follows;

P(C/R) = 1 - P(C/A)

∴ P(C/R) ≈ 1 - 0.99776 = 0.00224 ≈ 0.2%

The probability that a correctly weighed package is rejected, P(C/R) ≈ 0.2%

5 0
2 years ago
The table lists recommended amounts of food to order for 25 party guests. Amanda and Syndey are hosting a graduation party for 4
yan [13]

Answer:

Answer: 25 guests is 2.5 times more than 10

Multiply each order by 2.5

Chicken = 16 x 2.5 = 40

Lasagna = 7 x 2.5 = 17.5 , round to 18

Deli meat = 1.8 x 2.5 = 4.5, round to 5

Sliced cheese = 15 x 2.5 = 37.5, round to 38

Buns = 1 x 2.5 = 2.5, round to 3

Potato salad = 2 x 2.5 = 5

Step-by-step explanation:

7 0
2 years ago
Unit 4: Lesson 9: Parallel and Perpendicular Lines Unit Test Parallel and Perpendicular Lines does anyone have the answers for t
Nookie1986 [14]
What test is this question from?
7 0
2 years ago
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