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uysha [10]
3 years ago
7

Jill made scarves and then sold them for $15 each. If Jill spent $150 to make all the scarves, how many scarves need to be sold

so that Jill doesn't lose money?​
Mathematics
2 answers:
hram777 [196]3 years ago
7 0

Answer:

10 scarves

Step-by-step explanation:

lesantik [10]3 years ago
3 0

Answer:

10 scarves

Step-by-step explanation:

each scarf is $15, 10 of them is 150 because 10 x 15 = 150

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Maureen went on a 3 day 50 mile trip. the first day she biked 17 7/8 miles . the second day she biked 18 5/7 miles. How many mil
Inessa [10]
To find the miles for 3rd day, we take away the miles she biked for the first two day from the total.

   50 - 17 \dfrac{7}{8} - 18\dfrac{5}{7}

=50 - 17 \dfrac{7 \times 7}{8 \times 7} - 18\dfrac{5 \times 8}{7 \times 8}

= 48 \dfrac{112}{56}  - 17 \dfrac{49}{56} - 18\dfrac{40}{56}

= 13 \dfrac{23}{56}   \text { miles }

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Answer: She traveled 13 23/56 miles on the third day.
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3 years ago
Factor completely 5a2 + b.<br><br> Prime<br><br> a(5a + b)<br><br> b(5a2)<br><br> ab(5a + 1)
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\sf {5a^2+b~is~\boxed{\bf{prime}}}

It is prime because it has no common factors in 5a^2 and b.

The answer is A<span>.</span>
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What is the expanded form for 114,006 square miles
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This is the answer in expanded form: 100,000 + 10,000 + 4,000 + 6<span />
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26. Simplify the expression below completely.<br> 4(5w - 3) -(w - 1)<br> Please help me!!
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A researcher wishes to estimate the number of households with two cars. How large a sample is needed in order to be 99% confiden
solmaris [256]

Answer:

Sample size n = 1382

so correct option is D) 1382

Step-by-step explanation:

given data

confidence level = 99 %

margin of error = 3%

probability = 25 %

to find out

How large a sample size needed

solution

we know here P = 25 %

so 1 - P = 1 - 0.25

1 - P  = 0.75

and we know E margin of error is 0.03 so value of Z for 99%

α = 1 - 99%   = 1 - 0.99

α  = 0.01

and  \frac{\alpha}{2} = \frac{0.01}{2}

\frac{\alpha}{2}  = 0.005

so Z is here

Z_(\frac{\alpha}{2}) = 2.576

so

sample size will be

Sample size n =  (\frac{(Z_(\frac{\alpha}{2})}{E})^2 * P * (1-P)

put here value

Sample size n = (\frac{2.576}{0.03})^2 * 0.25 * 0.75

Sample size n = 1382

so correct option is D) 1382

3 0
3 years ago
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