The value of the 7 in 3,796.41 is of the value of the 7 in which of the following numbers?

Alfonso scored 1818 points during his last basketball game, which was 40\%40% of the points the entire team scored. The entire t

soldi70 [24.7K]

Answer: 45 points

Step-by-step explanation:

Let the point of the entire team be represented by x.

Alfonso scored 18 points during his last basketball game, which was 40% of the points the entire team scored. This can then be represented in an equation as:

40% of x = 18

40% × x = 18

40/100 × x = 18

0.4 × x = 18

0.4x = 18

x = 18/0.4

x = 45

The team's total point is 45

8 0

3 years ago

Which do you pick and why? How much money do you end up with in each case?

Mademuasel [1]

Answer:

1. Start with $1 and then double the money you have everyday for 30 days. You would end up with 1,073,741,824 on the 30th day.

Step-by-step explanation:

Why you should choose the first option:

1 x 2 = 2

2 x 2 = 4

4 x 2 = 8

8 x 2 = 16

16 x 2 = 32

32 x 2 = 64

64 x 2 = 128

128 x 2 = 256

256 x 2 = 512

512 x 2 = 1024

1024 x 2 = 2048

2048 x 2 = 4096

4096 x 2 = 8192

8192 x 2 = 16384

16384 x 2 = 32768

32768 x 2 = 65536

65536 x 2 = 131072

131072 x 2 = 262144

262144 x 2 = 524288

524288 x 2 = 1048576

1048576 x 2 = 2097152

2097152 x 2 = 4194304

4194304 x 2 = 8388608

8388608 x 2 = 16777216

16777216 x 2 = 33554432

33554432 x 2 = 67108864

67108864 x 2 = 134217728

134217728 x 2 = 268435456

268435456 x 2 = 536870912

536870912 x 2 = 1073741824

7 0

2 years ago

Damon borrowed money from his grandmother 3 years ago and agreed to pay her 4% simple annual interest. At the end of 3 years, he

sergeinik [125]

Answer:

200

Step-by-step explanation:

200×4%=8

8×3=24

The original amount is $200

6 0

3 years ago

Read 2 more answers

Choose the fraction that is equal to 0.245 A. 49/10 B. 245/999 C. 245/900 D. 245/1000

kicyunya [14]

D is equal to 0.245.

6 0

3 years ago

The process standard deviation is 0.27, and the process control is set at plus or minus one standard deviation. Units with weigh

mr_godi [17]

Answer:

a) $P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.159+0.159 = 0.318$

And the expected number of defective in a sample of 1000 units are:

$X= 0.318*1000= 318$

b) $P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.00135+0.00135 = 0.0027$

And the expected number of defective in a sample of 1000 units are:

$X= 0.0027*1000= 2.7$

c) For this case the advantage is that we have less items that will be classified as defective

Step-by-step explanation:

Assuming this complete question: "Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 10 ounces. Calculate the probability of a defect and the expected number of defects for a 1000-unit production run in the following situation.

Part a

The process standard deviation is .15, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects."

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(10,0.15)$

Where $\mu=10$ and $\sigma=0.15$

We can calculate the probability of being defective like this:

$P(X$

And we can use the z score formula given by:

$z=\frac{x-\mu}{\sigma}$

And if we replace we got:

$P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.159+0.159 = 0.318$

And the expected number of defective in a sample of 1000 units are:

$X= 0.318*1000= 318$

Part b

Through process design improvements, the process standard deviation can be reduced to .05. Assume the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects.

$P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.00135+0.00135 = 0.0027$

And the expected number of defective in a sample of 1000 units are:

$X= 0.0027*1000= 2.7$

Part c What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean?

For this case the advantage is that we have less items that will be classified as defective

5 0

3 years ago