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REY [17]
2 years ago
11

My sister and ate some of a chocolate bar she had 1/4 and I had 5/8 how much of the chocolate bar did we eat

Mathematics
2 answers:
Aleonysh [2.5K]2 years ago
8 0

Answer:

You ate 7/8 of the chocolate bar.

Step-by-step explanation:

First, we need to change 1/4 and 5/8 so that they have the same denominators (bottom numbers).

There's a common multiple of 8.

What you do to the bottom needs to be done to the top.

1 x 8 = 8

1 x 5 = 5

5/8

Your amount of chocolate stays the same.

4 x 2 = 8

1 x 2 = 2

2/8 is how much the sister had,

Finally, add the two fractions together.

2/8 + 5/8 = 7/8

You ate 7/8 of the chocolate bar.

Hope this helps, have a good day! :D

abruzzese [7]2 years ago
5 0

Answer:

7/8

Step-by-step explanation:

1/4 times 2/2 = 2/8. you would have to do this because the only way you can add fractions is if they have the same denominator. so 2/8 + 5/8= 7/8

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In a survey, 11 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped
pickupchik [31]

Answer:

The population standard deviation is not known.

90% Confidence interval by T₁₀-distribution: (38.3, 53.7).

Step-by-step explanation:

The "standard deviation" of $14 comes from a survey. In other words, the true population standard deviation is not known, and the $14 here is an estimate. Thus, find the confidence interval with the Student t-distribution. The sample size is 11. The degree of freedom is thus 11 - 1 = 10.

Start by finding 1/2 the width of this confidence interval. The confidence level of this interval is 90%. In other words, the area under the bell curve within this interval is 0.90. However, this curve is symmetric. As a result,

  • The area to the left of the lower end of the interval shall be 1/2 \cdot (1 - 0.90)= 0.05.
  • The area to the left of the upper end of the interval shall be 0.05 + 0.90 = 0.95.

Look up the t-score of the upper end on an inverse t-table. Focus on the entry with

  • a degree of freedom of 10, and
  • a cumulative probability of 0.95.

t \approx 1.812.

This value can also be found with technology.

The formula for 1/2 the width of a confidence interval where standard deviation is unknown (only an estimate) is:

\displaystyle t \cdot \frac{s_{n-1}}{\sqrt{n}},

where

  • t is the t-score at the upper end of the interval,
  • s_{n-1} is the unbiased estimate for the standard deviation, and
  • n is the sample size.

For this confidence interval:

  • t \approx 1.812,
  • s_{n-1} = 14, and
  • n = 11.

Hence the width of the 90% confidence interval is

\displaystyle 1.812 \times \frac{14}{\sqrt{10}} \approx 7.65.

The confidence interval is centered at the unbiased estimate of the population mean. The 90% confidence interval will be approximately:

(38.3, 53.7).

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