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Fynjy0 [20]
3 years ago
5

What is 4649+3383473

Mathematics
2 answers:
sp2606 [1]3 years ago
8 0

Answer:

3388122

Step-by-step explanation:

Crazy boy [7]3 years ago
5 0

Answer:

3388122

Step-by-step explanation:

just add them together or use a calculator

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9.8 m/s

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60% of the shakes at The Shake Shop have protein powder in them. There are 30 shakes on the menu. How many shakes have protein p
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Help pleaseeee- it’s for an assessment :/
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Step-by-step explanation:

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18. A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be
avanturin [10]

Answer:

a. =50 ± 4.67

b. =50 ± 4.81  

decreasing the sample size increase the margin of error

c. =50 ± 3.51

decreasing the confidence level reduced the margin of error

d. No.  because the confidence interval coefficient is gotten from the table of normal distribution of Z value.  therefore the confidence interval is only used for normal distributed population

Step-by-step explanation:

given

The sample mean, x, = 50,

the sample standard deviation, s, = 8.

a. Construct a 98% confidence interval for m if the sample size, n, is 20

for a 98% confidence interval of an infinite population, thus for a sample of size N and mean x, drawn from an infinite population having a standard deviation of σ, the mean value of the population is estimated to by:

x± 2.33σ /√N

for a confidence level of 98%, Zc = 2.33 gotten from the table of confidence interval.

This indicates that the mean value of the population lies between

50- ( 2.33*8 /√20)      and 50+ ( 2.33*8 /√20)

with 98% confidence in this prediction.

=50 ± 4.67

=45.33 or 54.67

(b) Construct a 98% confidence interval for m if the sample size, n, is 15.

using the same method as above

x± 2.33σ /√N

This indicates that the mean value of the population lies between

50- ( 2.33*8 /√15)      and 50+ ( 2.33*8 /√15)

=50 ± 4.81

=45.19 or 54.81

decreasing the sample size increase the margin of error

(c) Construct a 95% confidence interval for m if the sample size, n, is 20. Compare the results to those obtained in part

95% confidence interval =1.96 (gotten from table of confidence coefficient)

using x±1.96 σ /√N

This indicates that the mean value of the population lies between

50- (1.96 *8 /√20)      and 50+ (1.96 *8 /√20)

=50 ± 3.51

=46.49 or 53.51

c. decreasing the confidence level reduced the margin of error

(d) Could we have computed the confidence intervals in parts (a)–(c) if the population had not been normally distributed?

No.  because the confidence interval coefficient is gotten from the table of normal distribution of Z value.  therefore the confidence interval is only used for normal distributed population

7 0
3 years ago
LOOK AT THE PICTURE
OleMash [197]

Answer:

I don't think any of the answers here are right. They aren't a proportional relationship since they don't pass through the origin. So I would believe the answer would be C. Sorry if I'm wrong!!

Step-by-step explanation:

I learned about this stuff :D

6 0
3 years ago
Read 2 more answers
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