Laura has 10 1/2 yards of ribbon to make 7 bows.how much ribbon will she need to each bow

Mathematics

2 answers:

Ainat [17]3 years ago

7 0

10.5/7 is all you would need to do to find your answer of 1.5 yards of ribbon per bow.

Andreyy893 years ago

4 0

Well, we can simply divide 10 and 1/2 by 7, that's how many "yards" time 7 goes into 10 and 1/2.

$\bf \cfrac{\quad10\frac{1}{2} \quad }{7}\implies \cfrac{\quad \frac{10\cdot 2+1}{2}\quad }{\frac{7}{1}}\implies \cfrac{\quad \frac{21}{2}\quad }{\frac{7}{1}}\implies \cfrac{21}{2}\cdot \cfrac{1}{7}\implies \cfrac{3}{2}\cdot \cfrac{1}{1}
\\\\\\
\cfrac{3}{2}\implies \stackrel{yards}{1\frac{1}{2}}$

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3 years ago

The number of hours spent per week on household chores by all adults has a mean of 28 hours and a standard deviation of 7 hours.

Alchen [17]

Answer:

Probability that the mean hours spent per week on household chores by a sample of 49 adults will be more than 26.75 is 0.89435.

Step-by-step explanation:

We are given that the number of hours spent per week on household chores by all adults has a mean of 28 hours and a standard deviation of 7 hours.

Also, sample of 49 adults is given.

Let X = number of hours spent per week on household chores

So, assuming data follows normal distribution; X ~ N( $\mu=28,\sigma^{2}=7^{2}$ )

The z score probability distribution for sample mean is given by;

Z = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = mean hours = 28

$\sigma$ = standard deviation = 7 hours

n = sample of adults = 49

Let $\bar X$ = sample mean hours spent per week on household chores

So, probability that the mean hours spent per week on household chores by a sample of 49 adults will be more than 26.75 is given by =P( $\bar X$ > 26.75 hours)

P( $\bar X$ > 26.75 hours) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{26.75-28}{\frac{7}{\sqrt{49} } }$ ) = P(Z > -1.25) = P(Z < 1.25)