Two questions please help

What's the answer to this? and please explain.

vichka [17]

<h3>Answer: No, there isn't enough ribbon</h3>

===========================================================

Use this formula

a & b/c = (a*c+b)/c

to convert from a mixed number to an improper fraction

So,

a & b/c = (a*c+b)/c

3 & 3/4 = (3*4+3)/4

3 & 3/4 = 15/4

---------------------------------------------------

Next, we add the fractions 15/4 and 5/6. The denominators aren't the same, so we can't add right away. We need to get the denominators to the LCD.

In this case, the LCD is 4*6 = 24

The fraction 15/4 becomes 90/24 after multiplying top and bottom by 6.
The fraction 5/6 becomes 20/24 after multiplying top and bottom by 4

Now we can add the fractions:

90/24+20/24 = (90+20)/24 = 110/24

---------------------------------------------------

So far, we've added the 3&3/4 portion to the 5/6 portion to get the result 110/24

The last thing to do is to add on the two 1/3 ft pieces, which means we're adding on 2/3 of a foot

2/3 = 16/24 after multiplying top and bottom by 8

Add this to the result of the last section

110/24+16/24 = (110+16)/24 = 126/24

---------------------------------------------------

From here, we convert the improper fraction 126/24 to decimal form

Using a calculator, you should find that 126/24 = 5.25

The result is larger than 5, which means she used too much ribbon. There won't be enough ribbon to do everything she wants to do.

---------------------------------------------------

The shortcut we can take is to type the following into the calculator

3+3/4+5/6+2/3

doing so should lead to 5.25

5 0

3 years ago

Read 2 more answers

An instructor in a college class recently gave an exam that was worth a total of 100 points. The instructor inadvertently made t

solniwko [45]

Answer:

B) Method 2 will increase the standard deviation of the students' scores.

Step-by-step explanation:

Given that an instructor in a college class recently gave an exam that was worth a total of 100 points.

The average score for his students was 43 and the standard deviation of the scores was 5 points.

And now he is considering two different strategies for rescaling the exam results of which:

Method 1 = Add 17 points to everyone's score.

Method 2 = Multiply everyone's score by 1.7 .

And we have to check what will be the impact of these methods on the standard deviation of the students' scores.

For this let us consider a simple example to understand this:

Firstly, Formula for calculating Standard Deviation = $\sqrt{\frac{\sum (X-Xbar)^{2}}{n-1}}$

Suppose,

X X - Xbar $(X - Xbar)^{2}$

3 3 - 6 = -3 -3 * -3 = 9

5 5 - 6 = -1 -1 * -1 = 1

10 10 - 6 = 4 4 * 4 = 16

Mean of above data, Xbar = $\frac{3+ 5+10}{3}$ = 6

Standard Deviation of data = $\sqrt{\frac{26}{3-1} }$ = 3.6055

Now let us suppose that we multiply each value of above data with 2 so the new data will be:

X X - Xbar $(X - Xbar)^{2}$

3*2 = 6 6 - 12 = -6 -6 * -6 = 36

5*2 = 10 10 - 12 = -2 -2 * -2 = 4

10*2 =20 20 - 12 = 8 8 * 8 = 64

Mean of new data, Xbar = $\frac{6+ 10+20}{3}$ = 12

Standard Deviation of new data = $\sqrt{\frac{104}{3-1} }$ = 7.2111

Hence, we see that when we multiply any value to the data the standard deviation will increase and in other words it will multiplied by that value which value we multiplied with each data value i.e. when we multiply each data value with 2 the standard deviation also get multiplied by as

3.6055 * 2 = 7.2111

Therefore option B is correct that Method 2 will increase the standard deviation of the students' scores.

And on the other hand Similarly by adding any constant to the data the Standard Deviation will remain same. Therefore Method 1 will have no impact on standard deviation of the students' scores.

8 0

3 years ago

Georgia is making cupcakes at her bakery. she has enough batter to make 377 chocolate cupcakes, and 935 vanilla cupcakes. if she

olchik [2.2K]

All you have to do is add the number of vanilla and the number of chocolate, then divide the result by 24. You round up the result to the nearest whole number.

Like so:
377+935=1312
1312/24=54.67
55

So, she would have to refill it 55 times to use all the batter.

6 0

2 years ago

Read 2 more answers

A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben

juin [17]

Answer:

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

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

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.