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tresset_1 [31]
3 years ago
15

(30 POINTS!!!) One press can run the day’s newspapers in 6 hours, while another can do the same job in 8 hours. After running to

gether for 2 hours to complete the job, the faster one breaks. How much longer must the slower press run to finish the newspapers?
Mathematics
1 answer:
mr_godi [17]3 years ago
8 0

Amount of the job done after 2 hours:

2(1/6 + 1/8)

2(4/24 + 3/24)

2(7/24)

7/12 (amount of job finished)

.

amount of job left to do:

1 - 7/12

 

12/12 - 7/12

5/12 (remaining)

.

Let x = time (hours) slower press takes to finish job

then

x(1/6) = 5/12

multiplying both sides by 6:

x = 5/12 *6

x = 5/2 hours

or

x = 2 hours and 30 minutes

The slower press (8-hour press), will take h 3 hours, 20 minutes to complete the job

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What expression represents '' five times the quotient of some number and ten"? A.5(z/10) B.5(z-10) C.5(10/z) D.5(10-z)
maw [93]

Answer:

Option A is correct

5(\frac{z}{10})

Step-by-step explanation:

Let the number be z.

Given the statement:

'' five times the quotient of some number and ten"

" quotient of some number and ten" translated to \frac{z}{10}

"five times the quotient of some number and ten" translated to 5 \times \frac{z}{10}

Then, the expression we get,

5(\frac{z}{10})

Therefore, 5(\frac{z}{10}) expression represents '' five times the quotient of some number and ten"

5 0
3 years ago
Read 2 more answers
Given circle and circle with radii of 6cm and 4cm respectively.
noname [10]

Solution :

Given that :

The radius of circle A = 6 cm

The radius of circle C = 4 cm

In circle A

\angle EAF = \theta = 140^\circ

The length of arc EF = $2 \pi r \times \frac{\theta}{360^\circ}$

                                   $=2 \times 3.14 \times 6 \times \frac{140^\circ}{360^\circ}$

                                    = 14.653 cm

In circle C

\angle GCH = \theta = 140^\circ

The length of arc GH = $2 \pi r \times \frac{\theta}{360^\circ}$

                                   $=2 \times 3.14 \times 4 \times \frac{140^\circ}{360^\circ}$

                                    = 9.769 cm    

Therefore,

The length of EF is 14.653 cm

The length of GH is 9.769 cm

The length of EF is  1.5 times the length of GH

i.e.                   14.653  = 1.5 x 9.769

                       14.653 = 14.653

Hence proved.

                             

6 0
3 years ago
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
I am Lyosha [343]

Answer:

(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

Step-by-step explanation:

Let <em>X</em> = number of students who read above the eighth grade level.

(a)

A sample of <em>n</em> = 269 students are selected. Of these 269 students, <em>X</em> = 224 students who can read above the eighth grade level.

Compute the proportion of students who can read above the eighth grade level as follows:

\hat p=\frac{X}{n}=\frac{224}{269}=0.8327

The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

1-\hat p=1-0.8327

        =0.1673

Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

<em>n</em> = 709

<em>X</em> = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

\hat q=1-\hat p

  =1-\frac{X}{n}

  =1-\frac{546}{709}

  =0.2299\\\approx 0.229

The critical value of <em>z</em> for 95% confidence interval is:

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

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

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

     =0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)

Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

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lara [203]

Answer:

3

Step-by-step explanation:

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