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Viktor [21]
3 years ago
10

The original selling price of a jacket was s dollars. The selling price was then changed on two occasions by the store owner. It

s price is now represented by 0.85(1.4s). Which expression could explain what happened to the price of the jacket?​
Mathematics
1 answer:
solniwko [45]3 years ago
7 0

Answer:

B

Step-by-step explanation:

B) The price was increased by 40% and then decreased by 15%.

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Simplify this questions<br>6<br>---+15.4<br>5
sweet-ann [11.9K]
6/5 +15.4
12/10 +15.4
12/10+154/10
166/10
16.6

This can also be done by using the following method:
6/5 +15.4
12/10+15.4
1.2+15.4
16.6

Hope this helps :)
5 0
3 years ago
Jerome swam 11 laps in his circula pool. If he traveled a total distance of 1,036.2 feet, what is the diameter of his pool?
Cerrena [4.2K]

Answer:

5

Step-by-step explanation:

8 0
3 years ago
A magazine provided results from a poll of 500500 adults who were asked to identify their favorite pie. Among the 500500 ​respon
Komok [63]

Answer:

99% confidence interval is wider as compared to the 80% confidence interval.

Step-by-step explanation:

We are given that a magazine provided results from a poll of 500 adults who were asked to identify their favorite pie.

Among the 500 respondents, 14​% chose chocolate​ pie, and the margin of error was given as plus or minus ±3 percentage points. 

The pivotal quantity for the confidence interval for the population proportion is given by;

                            P.Q.  =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of adults who chose chocolate​ pie = 14%

            n = sample of adults = 500

            p = true proportion

Now, the 99% confidence interval for p =  \hat p \pm Z_(_\frac{\alpha}{2}_)  \times \sqrt{\frac{\hat p(1-\hat p)}{n} }

Here, \alpha = 1% so  (\frac{\alpha}{2}) = 0.5%. So, the critical value of z at 0.5% significance level is 2.5758.

Also, Margin of error = Z_(_\frac{\alpha}{2}_)  \times \sqrt{\frac{\hat p(1-\hat p)}{n} }  = 0.03 for 99% interval.

<u>So, 99% confidence interval for p</u>  =  0.14 \pm2.5758  \times \sqrt{\frac{0.14(1-0.14)}{500} }

                                                        = [0.14 - 0.03 , 0.14 + 0.03]

                                                        = [0.11 , 0.17]

Similarly, <u>80% confidence interval for p</u>  =  0.14 \pm 1.2816  \times \sqrt{\frac{0.14(1-0.14)}{500} }

Here, \alpha = 20% so  (\frac{\alpha}{2}) = 10%. So, the critical value of z at 10% significance level is 1.2816.

Also, Margin of error = Z_(_\frac{\alpha}{2}_)  \times \sqrt{\frac{\hat p(1-\hat p)}{n} }  = 0.02 for 80% interval.

So, <u>80% confidence interval for p</u>  =  [0.14 - 0.02 , 0.14 + 0.02]

                                                           =  [0.12 , 0.16]

Now, as we can clearly see that 99% confidence interval is wider as compared to 80% confidence interval. This is because more the confidence level wider is the confidence interval and we are more confident about true population parameter.

5 0
3 years ago
May I please get some help?
Vladimir79 [104]
I think its 80 m3 or either 18 m3
3 0
3 years ago
Read 2 more answers
Can someone help me please ? Thanks!
notka56 [123]
Use pythagorean theorem for all of them
a. 3, 4, 5
3^2+4^2?5^2
9+16?25
25=25
right 

b. 5, 6, 7
5^2+6^2=7^2
25+36=49
61>49
acute

c. 64+81=144
145>144
acute

hope this helps!
6 0
3 years ago
Read 2 more answers
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