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Anna [14]
3 years ago
11

Pls help ASAP I need help

Mathematics
2 answers:
iren [92.7K]3 years ago
8 0

Answer:

14

Step-by-step explanation:

94 - 10 = 84

84 / 6 = 14

Nesterboy [21]3 years ago
7 0
The value of x is 14 so the answer is 14 have a nice day ✌
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A cone has a volume of 120 pi cubic feet. If the radius is 3 feet, what
sergey [27]

Answer:

Step-by-step explanation:

V= 2pi^h

120=2(3,14) h

120/6,28= 19 feet

6 0
3 years ago
A multiple-choice test contains 10 questions. there are four possible answers for each question. (a) in how many ways can a stud
devlian [24]
A). 4^10 = 1,048,576 ways

b). 5^10 = 9,765,625 ways
5 0
3 years ago
By selling an article for 3600$, a man takes profit of 20%. What would his gain % be if he sold the article for 4,000$
ddd [48]

Answer:

28%

Step-by-step explanation:

3600$ × 20%

=3600 ×(20/100)

=3600 × 0.2

= 720

Actual Price = 3600 - 720

= 2880

Profit is at 4000$ = 4000 - 2880

= 1120$

%profit = (1120/4000) × 100

profit = 28%

3 0
4 years ago
Read 2 more answers
Construc t a 95% confidence interval for the population standard deviation σ of a random sample of 15 men who have a mean weight
GrogVix [38]

The 95% confidence interval for the population standard deviation will be,

$\sqrt{\frac{(n-1) s^{2}}{\chi^{2}} \frac{\alpha}{2}} & < \sigma < \sqrt{\frac{(n-1) s^{2}}{\chi^{2}}} \\

simplifying the equation, we get

$\sqrt{\frac{2551.5}{26.1}} & < \sigma < \sqrt{\frac{2551.5}{5.63}} \\9.887 & < \sigma < 21.288

The 95% confidence interval exists (9.887, 21.288).

<h3>What is the  95% of confidence interval?</h3>

A random sample of 15 men exists selected. The mean weight exists at $165.2 pounds and the standard deviation exists at $13.5 pounds. The population exists normally distributed.

So, $n=15, \bar{X}=165.2, s=13.5$

where n exists the sample size, $\bar{X}$ exists the sample size

s exists the sample standard deviation.

The degrees of freedom will be n - 1 i.e.15 - 1 = 14.

For a 95% confidence interval, the level of significance will be $\alpha=0.05$.

The 95% confidence interval for the population standard deviation will be,

$\sqrt{\frac{(n-1) s^{2}}{\chi^{2}} \frac{\alpha}{2}} & < \sigma < \sqrt{\frac{(n-1) s^{2}}{\chi^{2}}} \\

substitute the values in the above equation, we get

$\sqrt{\frac{(15-1)(13.5)^{2}}{\chi^{2}} 0.05} & < \sigma < \sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0}^{2}}} \\

$\sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0.975}^{2}}} & < \sigma < \sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0.025}^{2}}} \\

simplifying the above equation, we get

$\sqrt{\frac{2551.5}{26.1}} & < \sigma < \sqrt{\frac{2551.5}{5.63}} \\9.887 & < \sigma < 21.288

The 95% confidence interval exists (9.887, 21.288).

To learn more about confidence interval refer to:

brainly.com/question/14825274

#SPJ4

6 0
2 years ago
I need help with angles again.
zaharov [31]

Answer:

119 degrees

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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