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3 years ago

Name five fractions whose values are between 1/8 and 1/5

Mathematics

2 answers:

TiliK225 [7]3 years ago

8 0

The answer is 1/8 bc thats the anwser i got

Volgvan3 years ago

6 0

Answer:

1/5

1. 1/56

2. 1/6

3. 1/65

4. 1/7

5. 1/75

1/8

Step-by-step explanation:

The fractions I gave (1/56, 1/6, 1/65, 1/7, 1/75) are all in between 1/5 and 1/8.

A way to think about this is to add a zero to a denominator of 1/5 and 1/8 and turn them into whole numbers:

1/5 --> 1/50 --> 50

1/8 --> 1/80 --> 80

Then find number between them:

55, 60, 65, 70, 75

Then, turn them into a fraction with a 1 as the numerator and the number as denominators:

1/55, 1/60, 1/65, 1/70, 1/75

This is the process. These numbers are all in between 1/5 and 1/8.

I hope this helps you!

(P.S., I don't think this works with negative numbers, though)

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Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of

freedom are -1.796 & 1.796 with P = 5%}

P(-1.796 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.796) = 0.90

P( $-1.796 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }$ , $21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }$ ]

= [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0

3 years ago

I still can’t figure out this stupid simple interest question! I tried several answer n NONE worked so I’m gonna try 1 last time

natka813 [3]

t = $\frac{I}{Pr}$

isolate t by dividing both sides by Pr

$\frac{I}{Pr}$ = t ⇒ t = $\frac{I}{Pr}$

7 0

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