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patriot [66]
2 years ago
13

Question 1 of 10

Mathematics
1 answer:
sp2606 [1]2 years ago
3 0

20 raise the power 7

Step-by-step explanation:

Which clearly means 20 × itself 7 times

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What is 53 divided by 76
Goshia [24]

Answer:

0.69

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Two angle measures of a quadrilateral are 34° and 66°. What could the measure of the other two angles be?
alexdok [17]
Can u upload the diagram if u can
3 0
3 years ago
Write a cosine function that has a midline of 2, an amplitude of 4 and a period of pi/2.
exis [7]

y = Acos(Bx) + D;

D = 4, A = 2. Now T = 2π/B = 5π/8, B = 2π/(5π/8) = 16/5

WE get y = 2cos(16/5x) + 4

4 0
2 years ago
Hey I’m struggling with number 13 so please show your work. Thx
Dafna11 [192]

Answer:

\frac{5}{12}

Step-by-step explanation:

Because \frac{3}{5} is being multiplied by S, we can divide \frac{3}{5} from both sides of the equation. This will give us:

\frac{1}{4}÷\frac{3}{5}

But, that looks a bit hectic. Instead of dividing, you can multiply by the reciprocal (which is essentially how to divide fractions). So, instead of \frac{1}{4}÷\frac{3}{5}, you get:

\frac{1}{4}×\frac{5}{3}

When multiplying fractions, remember you can just multiply straight across-- numerator x numerator, then denominator x denominator.

By doing that, you get the fraction \frac{5}{12}

\frac{5}{12} cannot be simplified any more, so S=\frac{5}{12} is your answer :)

I hope this helps

6 0
3 years ago
A software developer wants to know how many new computer games people buy each year. Assume a previous study found the standard
Marrrta [24]

Answer:

The minimum sample size required to ensure that the estimate has an error of at most 0.14 at the 95% level of confidence is n=567.

Step-by-step explanation:

We have to calculate the minimum sample size n needed to have a margin of error below 0.14.

The critical value of z for a 95% confidence interval is z=1.96.

To do that, we use the margin of error formula in function of n:

MOE=\dfrac{z\cdot \sigma}{\sqrt{n}}\\\\\\n=\left(\dfrac{z\cdot \sigma}{MOE}\right)^2=\left(\dfrac{1.96\cdot 1.7}{0.14}\right)^2=(23.8)^2=566.42\approx 567

The minimum sample size to have this margin of error is n = 567.

3 0
3 years ago
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