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MrRissso
8 days ago
15

Ruth has 12 feet of ribbon.

Mathematics
1 answer:
tester [8.8K]8 days ago
6 0

Response:

I’m unsure of the correct answer, though I solved a similar problem; feel free to report me if that doesn’t assist you.

Detailed explanation:

You can tackle this issue in two ways, both yielding the same result.

To visualize it better, it may help to convert to a more manageable unit.

Given your ribbon measures 5 feet long, divided into 6 parts, you know each won't exceed 1 foot in length.

One method involves converting your 5-foot ribbon into inches for easier calculations.

1 foot equals 12 inches

To determine the ribbon's length in inches, multiply its feet by 12.

Once acquired, you can divide by 6 to ascertain each segment's inch length.

Be cautious! The outcome must be converted back to feet; divide by 12 to get the final answer in feet, rounding to the nearest tenth.

Alternately, you might simply divide the 5 feet by 6 and round to the nearest tenth. This may be less clear but will yield the same outcome as converting to inches and back to feet.

Best of luck! I hope this proves useful!

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David is performing the following construction. Based on the markings that are present, what construction is he performing?
Svet_ta [9500]

Answer:

He is dividing the angle BAC into two equal parts.

Step-by-step explanation:

Initially, he places the compass at point A and draws two small arcs intersecting points D and E. Next, setting the compass at D and then at E, he draws two arcs that intersect between the line segments AB and AC.

The bisecting line is drawn from point A through the intersection of these arcs.

8 0
1 month ago
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Each leg of a 45°-45°-90° triangle measures 12 cm. Triangle X Y Z is shown. Angle X Y Z is a right angle and angles Y Z X and Z
babunello [8412]
Using the Pythagorean theorem: In a right triangle where legs are labeled a and b, and hypotenuse is c. Given the triangle ΔXYZ has a right angle at Y, with angles YZX and ZXY both measuring 45 degrees, and ZY = YX = 12 cm. The task is to find the length of XZ.
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10 days ago
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I helped 3.6 customer per half hour yesterday. so i helped _____ customers in 7.5 hour yesterday
lawyer [9240]

Customers served in 0.5 hours : 3.6   versus   x customers served in 7.5 hours

--------------------   = --------------------

0.5 hours                     7.5 hours



Applying cross multiplication:

3.6 multiplied by 7.5 equals x times 0.5

Dividing both sides by 0.5:

x = (3.6 * 7.5) / 0.5

x = 54

So, you assisted 54 customers in 7.5 hours.

7 0
1 month ago
Twelve of the 20 students in Mr. Skinner’s class brought lunch from home. Fourteen of the 21 students in Ms. Cho’s class brought
Svet_ta [9500]

Solution:

In Mr. Skinner's class, the count of students bringing lunch from home is 12 out of 20.

Fraction of students who brought lunch from home in Mr. Skinner's class=\frac{12}{20}=\frac{3}{5}

For Ms. Cho's class, the number who brought lunch from home is 14 out of 21.

Fraction of students who brought lunch from home in Ms. Cho's class=\frac{14}{21}=\frac{2}{3}

Siloni is utilizing two spinners with 15 equal sections to randomly select students from the classes and predict whether they brought lunch or will purchase it from the cafeteria.

Number of Equal sections in each Spinner=15

To visualize the students from Mr. Skinner's class who brought lunch using a Spinner with 15 equal sections =\frac{9}{15}

For Ms. Cho's class, using a Spinner with 15 equal sections =\frac{10}{15}

Mr. Skinner's Class +1 = Ms. Cho's Class

This means that the spinner for Ms. Cho's class will include one additional section representing students who brought lunch.

Option A signifies that one additional section on Mr. Skinner's spinner represents students who brought lunch, reflecting Ms. Cho's class.

9 0
26 days ago
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Question 1 Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your metho
babunello [8412]

Answer:

(a) The 95% confidence interval representing the percentage of the entire U.S. population that would select American football as their preferred television sport is (0.34, 0.40).

(b) Not reasonable.

Step-by-step explanation:

Given:

n = 1000

\hat p = 0.37

(a)

The confidence interval (1 - α)% for proportion p is:

CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

Where:

\hat p = the sample proportion

n = sample size

z_{\alpha/2} = critical z value.

Calculate the critical value of z for a 95% confidence level as shown:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Refer to a z-table for the necessary value.

Compute the 95% confidence interval for proportion p as follows:

CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

=0.37\pm 1.96\times\sqrt{\frac{0.37(1-0.37)}{1000}}

=0.37\pm 0.03\\=(0.34, 0.40)

Thus, the 95% confidence interval indicating the proportion of individuals in the U.S. who might say their preferred sport on TV is American football is (0.34, 0.40).

(b)

Next, we must assess if it's rational to consider that the actual percent of people in the United States who favor American football on television is 33%.

The hypothesis is defined as:

H₀: The portion of the U.S. population claiming American football as their favorite sport on television is 33%, meaning p = 0.33.

Hₐ: The proportion of people in the U.S. whose favorite sport to watch on television is not 33%, or p ≠ 0.33

This hypothesis can be verified using a confidence interval.

The decision rule:

If the (1 - α)% confidence interval contains the null value of the hypothesis, then the null hypothesis is not rejected. If, however, the (1 - α)% confidence interval excludes the null value of the hypothesis, then the null hypothesis is rejected.

<pthe confidence="" interval="" for="" the="" proportion="" of="" all="" u.s.="" individuals="" indicating="" that="" american="" football="" is="" their="" favorite="" sport="" on="" television="">

The confidence interval does encompass the null value of p, which is 0.33.

<pthus the="" null="" hypothesis="" will="" be="" rejected.=""><pin conclusion="" it="" is="">not reasonable to accept that 33% represents the actual percentage of those in the U.S. whose favorite televised sport is American football.

</pin></pthus></pthe>
6 0
1 month ago
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