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3 years ago

How do you solve this problem and what did you do to gain the answer 1/64+5/8-3/32=?

2 answers:

3 0

The answer is 35/64 or 0.546875 in decimal form

first we write all numerators above the least common denominator 64
1+40-6
————
64

next we solve the numerators by doing 1+40-6 to get 35/64

LUCKY_DIMON [66]3 years ago

3 0

Answer:

the answer is 35/64(in fraction) but in decimals it's 0.55

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After serving out 24 gallons, a milk tank had 24 pints left. How much milk was in the tank to begin with?

mel-nik [20]

Answer:

216 pints or 27 gallons

Step-by-step explanation:

Basically you have to convert the gallons into pints to make it easier to add.

1 gallon = 8 pints

To turn 24 gallons into pints, multiply 24*8= 192.

So, now you just add 24+192 which is 216.

To convert into gallons, divide by 8 so it is 27.

216 pints or 27 gallons

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3 years ago

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3 years ago

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Which is true of dependent events?

insens350 [35]

Answer:

Might be

Step-by-step explanation:

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5 0

3 years ago

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

2/5 - 1/5

hram777 [196]

All the fractions that you are subtracting have the same denominator so you just subtract the numerator.2/5-1/5=1/5
3/8-2/8=1/8
79-3/9 since you can simplify 3/9 by 3 you get 1/3 you get 79/1-1/3.the common denominator is 3 so you multiply 79/1 by 3 and 1/3 by 1 and then you get 237/3-1/3 and you get 236/3. then you turn it from an improper fraction into a mixed number. 236 divided by and you get 78.6 repeating

8 0

3 years ago