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3 years ago

Plz help with this math problem plz

Mathematics

2 answers:

3241004551 [841]3 years ago

7 0

Answer:

what problems. I could't see

erastova [34]3 years ago

6 0

Please upload a picture of your problem!

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One-third cup of black beans provides 20% of the potassium you need daily. You must get the remaining 1,360 milligrams from othe

VARVARA [1.3K]

5×20%=100% so 1360mg is 80% of the daily requirement. The total requirement is 1360/0.8=13600/8=1700 mg.

6 0

3 years ago

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G−9=−19 Help plz Pleaseeeeee

Kay [80]

Answer:

g = -10

Step-by-step explanation:

5 0

3 years ago

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Use a graphing calculator to solve the equation -2sin(theta)= 1.2 in the interval from 0 to 2pi.

elena-s [515]

Answer:

$\theta=\pi +arcsin(\frac{3}{5})\approx 3.785\\\\\theta=2\pi -arcsin(\frac{3}{5} ) \approx 5.64$

Step-by-step explanation:

As you can see from the graph I attached you, the possible solutions in the interval from 0 to 2π are approximately:

$\theta=3.7\\\\and\\\\\theta=5.7$

So, it's useful to solve the equation too, in order to verify the result:

$-2sin(\theta)=1.2\\\\sin(\theta)=-\frac{3}{5}$

Taking the inverse sine of both sides:

$\theta=\pi +arcsin(\frac{3}{5})\approx 3.785\\\\\theta=2\pi -arcsin(\frac{3}{5} ) \approx 5.64$

Using this result we can conclude the solutions in the interval from 0 to 2π are approximately:

$\theta\approx 3.785\\\\\theta \approx 5.64$

8 0

3 years ago

A college student organization wants to start a nightclub for students under the age of 21. To assess support for this proposal,

eduard

Answer:

a) Sample size = 822

b) Margin of error = 0.03407

Step-by-step explanation:

We are given the following in the question:

p = 74% = 0.74

a) Sample size is required to obtain margin of error of 0.03

Formula:

$\text{Margin of error} = z_{\text{statistic}}\times \sqrt{\dfrac{p(1-p)}{n}}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting values, we get,

$0.03 = 1.96\times \sqrt{\dfrac{0.74(1-0.74)}{n}}\\\\n = (\dfrac{1.96}{0.03})^2(0.74)(1-0.74)\\\\n = 821.24 \approx 822$

Thus, the sample size must be approximately 822 to obtain a 95% confidence interval with an approximate margin of error of 0.03

b) Margin of error for the 95% confidence interval

p = 54% = 0.54

Formula:

$\text{Margin of error} = z_{\text{statistic}}\times \sqrt{\dfrac{p(1-p)}{n}}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting values, we get,

$\text{Margin of error} = 1.96\times \sqrt{\dfrac{0.54(1-0.54)}{822}}\\\\=0.03407$

The margin of error now will be 0.03407.

5 0

3 years ago

Five students came for after-school tutoring. Lin’s teacher assigned each of them the same number of problems to complete. Then

horsena [70]

Answer:

problems to complete. Then he assigned each student 2 more problems. 30 problems were assigned in

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2. Five students came for after-school tutoring. Priya’s teacher assigned each of them the same number of

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3 years ago

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