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fredd [130]
3 years ago
14

What’s is the answer

Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
8 0

Answer:

In simplest form, it should be:

3:2

6 / 2 = 3

4 / 2 = 2

Hope this helps!


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a worker can complete the assembly of 15 chairs in 6 hours. At the rate,how many can the worker complete in a 40-hour work week?
san4es73 [151]

Answer:

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Step-by-step explanation:

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3 years ago
Karen already owns 3 hair bands, and additional hair bands are priced at 8 for a dollar. How much money does Karen need to spend
kiruha [24]

Answer:

Karen needs to spend $4.5 to buy new hair bands.

Step-by-step explanation:

Given:

Hair bands Karen have = 3

cost of 8 hair bands = $1

Total Hair bands she require = 39

Hence Number of of new hair bands require = Total Hair bands she require - Hair bands Karen have = 39 -3 = 36

we need to find how much money she need to spend to buy new hair bands.

cost of 8 hair bands = $1

Cost of 36 hair bands = money need to spend on hair bands

By using Unitary method we get;

Money spend on new hair bands = \frac{36}{8} = \$4.5

Hence,Karen needs to spend $4.5 to buy new hair bands.

8 0
3 years ago
11p-4=6p+1 explain steps please ASAP!!!
KatRina [158]
Attached below is the problem that I worked out. And I checked my answer simply by plugging it in

8 0
3 years ago
50 points!! ( don't answer if you don't know! WIll report) I REALLY NEED HELP!
larisa86 [58]

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Step-by-step explanation:

5 0
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Read 2 more answers
In a random sample of students who took the SAT test, 427 had paid for coaching courses and the remaining 2733 had not. Calculat
Dvinal [7]

Answer:

The 95% confidence interval for the proportion of students who get coaching on the SAT is (0.1232, 0.147).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

427 had paid for coaching courses and the remaining 2733 had not.

This means that n = 427 + 2733 = 3160, \pi = \frac{427}{3160} = 0.1351

95% confidence level

So \alpha = 0.05, z is the value of Z that has a p-value of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1351 - 1.96\sqrt{\frac{0.1351*0.8649}{3160}} = 0.1232

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1351 + 1.96\sqrt{\frac{0.1351*0.8649}{3160}} = 0.147

The 95% confidence interval for the proportion of students who get coaching on the SAT is (0.1232, 0.147).

8 0
3 years ago
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