Ask question

Ask question

All categories

2 years ago

7

Jalisa is using 5/6 yard of ribbon to make bows for her party favors. Jalisa needs to make 6 bows. What is the length of the rib

bon used for each bow?

2 answers:

Alex787 [66]2 years ago

5 0

Answer: The required length of the ribbon used for each bow is 0.12 yard.

Step-by-step explanation: Given that Jalisa is using $\dfrac{5}{6}$ yard of ribbon to make bows for her party favors and Jalisa needs to make 6 bows.

We are to find the length of the ribbon used for each bow.

We will be using the UNITARY method to solve the problem.

Length of the ribbon used to make 6 bows is

$\dfrac{5}{6}~\textup{yard}.$

So, the length of the ribbon used to make one row is given by

$\dfrac{\frac{5}{6}}{6}=\dfrac{5}{6}\times\dfrac{1}{6}=\dfrac{5}{36}=0.12.$

Thus, the required length of the ribbon used for each bow is 0.12 yard.

Leya [2.2K]2 years ago

4 0

5/36 yard. 5/6 and multiply by the reciprocal of 6, 1/6.

You might be interested in

Suppose you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by model

Ann [662]

Answer:

(1) The degrees of freedom for unequal variance test is (14, 11).

(2) The decision rule for the 0.01 significance level is;

If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.

If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.

(3) The value of the test statistic is 0.3796.

Step-by-step explanation:

We are given that you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by models featuring Liz Claiborne's attire with those of Calvin Klein.

The following is the amount ($000) earned per month by a sample of 15 Claiborne models;

$3.5, $5.1, $5.2, $3.6, $5.0, $3.4, $5.3, $6.5, $4.8, $6.3, $5.8, $4.5, $6.3, $4.9, $4.2 .

The following is the amount ($000) earned by a sample of 12 Klein models;

$4.1, $2.5, $1.2, $3.5, $5.1, $2.3, $6.1, $1.2, $1.5, $1.3, $1.8, $2.1.

(1) As we know that for the unequal variance test, we use F-test. The degrees of freedom for the F-test is given by;

$\text{F}_(_n__1-1, n_2-1_)$

Here, $n_1$ = sample of 15 Claiborne models

$n_2$ = sample of 12 Klein models

So, the degrees of freedom = ( $n_1-1, n_2-1$ ) = (15 - 1, 12 - 1) = (14, 11)

(2) The decision rule for 0.01 significance level is given by;

If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.

If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.

(3) The test statistics that will be used here is F-test which is given by;

T.S. = $\frac{s_1^{2} }{s_2^{2} } \times \frac{\sigma_2^{2} }{\sigma_1^{2} }$ ~ $\text{F}_(_n__1-1, n_2-1_)$

where, $s_1^{2}$ = sample variance of the Claiborne models data = $\frac{\sum (X_i-\bar X)^{2} }{n_1-1}$ = 1.007

$s_2^{2}$ = sample variance of the Klein models data = $\frac{\sum (X_i-\bar X)^{2} }{n_2-1}$ = 2.653

So, the test statistics = $\frac{1.007}{2.653 } \times 1$ ~ $\text{F}_(_1_4,_1_1_)$

= 0.3796

Hence, the value of the test statistic is 0.3796.

3 0

2 years ago

There are 8 sophomores on the academic team. At the last competition, they each took the math test. Their scores were 82%, 92%,

Sav [38]

Answer:

79%

Step-by-step explanation:

To find the median, you need to arrange the numbers in ascending order. Than pick the middle number. If there are two middle numbers then you add them together and then divide by 2.

3 0

2 years ago

Which of the following expressions is equivalent to the expression 5(4n - 11) - 3n + 20?

WINSTONCH [101]

17n-35 would be the correct answer

7 0

2 years ago

Read 2 more answers

Of the 500 sample households in the previous exercise, 7 had three or more large-screen TVs. (a) The percentage of households in

likoan [24]

Answer:

a) The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

$\hat p=\frac{7}{500}=0.014$

And that represent the 1.4%.

b) And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

Step-by-step explanation:

Data given and notation

n=500 represent the random sample taken

X=7 represent the households with three or more large-screen TVs

$\hat p=\frac{7}{500}=0.014$ estimated proportion of households with three or more large-screen TVs

$\alpha=0.05$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

p= population proportion of households with three or more large-screen TVs

Part a

The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

$\hat p=\frac{7}{500}=0.014$

And that represent the 1.4%.

Part b

Yes is possible. We hav that $np>10$ and $n(1-p)>10$ so we have the assumption of normality to find the interval.

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.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.014 - 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.00370$

$0.014 + 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.0243$

And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

4 0

2 years ago

In a group of 100 students, 25 own a laptop, 40 own a mobile phone, and 35 own either a

Talja [164]

Answer:

My best answer is that the probability is 35/100 that the student chosen at random owns neither a laptop nor a mobile phone.

Step-by-step explanation:

If you go back and read the introductory passage, you will notice it says, "In a group of 100 students... 35 own either a laptop or a mobile phone, but not both." And I think the other numbers, I.E. 25 and 40, were just meant to throw us off. Therefore, I think 35/100 is the probability.

8 0

2 years ago