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2 years ago

How do I do this next two problems???

Mathematics

1 answer:

rosijanka [135]2 years ago

8 0

Donuts cost 80 cents each and cup of coffee cost 1.25$

Do you need evidence?

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70% in the SIMPLEST Form

masya89 [10]

Answer:

7/100

Step-by-step explanation:

Find the GCD (or HCF) of numerator and denominator

GCD of 70 and 100 is 10

Divide both the numerator and denominator by the GCD

70 ÷ 10

100 ÷ 10

Reduced fraction:

Therefore, 70/100 simplified to lowest terms is 7/10.

4 0

3 years ago

Help me please! Thanks

Sunny_sXe [5.5K]

Hope this helps! have a nice day/night

8 0

2 years ago

In ΔMNO, m = 50 cm, o = 35 cm and ∠O=83°. Find all possible values of ∠M, to the nearest degree.

Alexxandr [17]

Answer:

No solution

Step-by-step explanation:

I got it right on delta math, just dont write anything and turn that in and it should be correct.

7 0

2 years ago

Four friends went scuba diving today. Ali dove 70 feet, Tim went down 50 feet, Carl dove 65 feet, and Brenda reached 48 feet bel

ss7ja [257]

Ali dove the furthest

7 0

3 years ago

Assume we need to estimate the mean of a normally-distributed population with great accuracy. Specifically, for significance lev

bulgar [2K]

Answer:

$n = 2662.56\sigma^2$

Step-by-step explanation:

We are given the following in the question:

Significance level = 0.01

Width of interval = 0.1

Population variance = $\sigma^2$

We have to find the sample size so that the width of the confidence interval is no larger than 0.1

$z_{critical}\text{ at}~\alpha_{0.01} = \pm 2.58$

Formula for sample size:

$n = \displaystyle\frac{z^2\sigma^2}{E^2}$

where E is the margin of error. Since the confidence interval width is 0.1,

$E = 0.05$

Putting these values in the equation:

$n = \displaystyle\frac{(2.58)^2\sigma^2}{(0.05)^2} = 2662.56\sigma^2$

So, the above expression helps us to calculate the sample size so that the width of the confidence interval is no larger than 0.1 for different sample variances.

4 0

3 years ago