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nata0808 [166]
3 years ago
10

Please help if so thank you!

Mathematics
2 answers:
katovenus [111]3 years ago
8 0

In order to get the answer for this question you need to times 2 by 10, which equals 20 since it's $2 for each guest. That now leaves her with $20. $26+$20 <u>equals </u><u>$46</u> for the total cost of the retirement party.

Snezhnost [94]3 years ago
7 0

Forgive me if im wrong.

I think the answer is 46.


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What is the r-value of the following data to three decimal places?
allsm [11]

Answer: it’s A

Step-by-step explanation:

7 0
3 years ago
F (x)<br> + 12 and g(x)=2x, what is the value of (1 = g)(144)?<br> 84<br> 60<br> 0<br> 48
Troyanec [42]
The question that I’ve typed is hard to read can type it again or take a picture
4 0
3 years ago
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
Can someone please help me with this?
horrorfan [7]
Um this seems really confusing!
3 0
4 years ago
39/51 change to lowest terms
OLEGan [10]

Answer:

As a proper fraction  - 39/51 = 13/17

As a decimal number:  39/51 ≈ 0.76

As a percentage:  39/51 ≈ 76.47%

Step-by-step explanation:

Calculate the greatest (highest) common factor (divisor), gcf (gcd).

13/17 =

13 ÷ 17 =

0.764705882353 ≈

0.76

By precentage

0.764705882353 =

0.764705882353 × 100/100 =

76.470588235294/100 =

76.470588235294% ≈

76.47%

7 0
3 years ago
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