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Nadya [2.5K]
3 years ago
13

Write one hundred and fourteen thousandths as a decimal number.

Mathematics
1 answer:
LiRa [457]3 years ago
6 0

100.014. I think this is it

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Need help on this... Thabks
Mashcka [7]
5(x+40-7) < 23

5x +33 < 23

5x < 23 -33

5x < -10

x < 2

Hope it helps I couldn’t do the line under the less than sign but I hope it helps!
3 0
3 years ago
Jill’s bowling scores are approximately normally distributed with mean 170 and standard deviation 20, while Jack’s scores are ap
miss Akunina [59]

Answer:

a) The probability of Jack scoring higher is 0.3446

b) They probability of them scoring above 350 is 0.2119

Step-by-step explanation:

Lets call X the random variable that determines Jill's bowling score and Y the random variable that determines jack's. We have

X \simeq N(170,400)\\Y \simeq N(160,225)

Note that we are considering the variance on the second entry, the square of the standard deviation.

If we have two independent Normal distributed random variables, then their sum is also normally distributed. If fact, we have this formulas:

N(\lambda_1, \sigma^2_1) + N(\lambda_2, \sigma^2_2) = N(\lambda_1 + \lambda_2,\sigma^2_1 + \sigma^2_2) \\r* N(\lambda_1, \sigma^2_1) = N(r\lambda_1,r^2\sigma^2_1)  

for independent distributions N(\lambda_1, \sigma^2_1) , N(\lambda_2, \sigma^2_2) , and a real number r.

a) We define Z to be Y-X. We want to know the probability of Z being greater than 0. We have

Z = Y-X = N(160,225) - N(170,400) = N(160,225) + (N(-170,(-1)^2 * 400) = N(-10,625)

So Z is a normal random variable with mean equal to -10 and vriance equal to 625. The standard deviation of Z is √625 = 25.

Lets work with the standarization of Z, which we will call W. W = (Z-\mu)/\sigma = (Z+10)/25. W has Normal distribution with mean 0 and standard deviation 1. We have

P(Z > 0) = P( (Z+10)/25 > (0+10)/25) = P(W > 0.4)

To calculate that, we will use the <em>known </em>values of the cummulative distribution function Φ of the standard normal distribution. For a real number k, P(W < k) = Φ(k). You can find those values in the Pdf I appended below.

Since Φ is a cummulative distribution function, we have P(W > 0.4) = 1- Φ(0.4)

That value of Φ(0.4) can be obtained by looking at the table, it is 0.6554. Therefore P(W > 0.4) = 1-0.6554 = 0.3446

As a result, The probability of Jack's score being higher is 0.3446. As you may expect, since Jack is expected to score less that Jill, the probability of him scoring higher is lesser than 0.5.

b) Now we define Z to be X+Y Since X and Y are independent Normal variables with mean 160 and 170 respectively, then Z has mean 330. And the variance of Z is equal to the sum of the variances of X and Y, that is, 625. Hence Z is Normally distributed with mean 330 and standard deviation rqual to 25 (the square root of 625).

We want to know the probability of Z being greater that 350, for that we standarized Z. We call W the standarization. W is s standard normal distributed random variable, and it is obtained from Z by removing its mean 330 and dividing by its standard deviation 25.

P(Z > 350) = P((Z  - 330)/25 > (350-330)/25) = P(W > 0.8) = 1-Φ(0.8)

The last equality comes from the fact that Φ is a cummulative distribution function. The value of Φ(0.8) by looking at the table is 0.7881, therefore P(X+Y > 350) = 1 - Φ(0.8) = 0.2119.

As you may expect, this probability is pretty low because the mean value of the sum of their combined scores is quite below 350.

I hope this works for you!

Download pdf
6 0
3 years ago
A small business ships specialty homemade candies to anywhere in the world. Past records indicate that the weight of orders is n
Fantom [35]

Answer:

90% confidence interval for the true mean weight of orders is between a lower limit of 103.8645 grams and an upper limit of 116.1355 grams.

Step-by-step explanation:

Confidence interval for true mean weight is given as sample mean +/- margin of error (E)

sample mean = 110 g

sample sd = 14 g

n = 16

degree of freedom = n - 1 = 16 - 1 = 15

confidence level = 90% = 0.9

significance level = 1 - C = 1 - 0.9 = 0.1 = 10%

critical value (t) corresponding to 15 degrees of freedom and 10% significance level is 1.753

E = t × sample sd/√n = 1.753×14/√16 = 6.1355 g

Lower limit of sample mean = sample mean - E = 110 - 6.1355 = 103.8645 g

Upper limit of sample mean = sample mean + E = 110 + 6.1355 = 116.1355 g

90% confidence interval is (103.8645, 116.1355)

4 0
3 years ago
Read 2 more answers
What is the scale factor from small triangle to the larger triangle
Lyrx [107]

Answer:

scale factor = 3

Step-by-step explanation:

The scale factor is the ratio of corresponding sides, larger to smaller.

scale factor = \frac{BC}{TB} = \frac{12}{4} = 3

7 0
2 years ago
Will mark barinly______
Lady_Fox [76]

Answer:

x = 13

y = 15

Step-by-step explanation:

ST = DT ( Reason : from given )

5x = 65

x = 65 / 5

x = 13

TY = TU ( Reason : from given )

3y + 5 = 50

3y = 50 - 5

3y = 45

y = 45 / 3

y = 15

8 0
3 years ago
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