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3 years ago

Priscilla can make 4 bracelets in 19 minutes. At this rate, how many bracelets can she make in 76 minutes?

Mathematics

1 answer:

irakobra [83]3 years ago

6 0

19 min = 4 bracelet
4 x 19 min = 76 min
If 19 min is 4 bracelets then 4 x 4 = 16 bracelets

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Trenton and Maria record how much dry food their pets eat on average each day.• Trenton's pet: 4/5 cup of dry food• Maria's pet:

Phoenix [80]

6.3 many more cups of dry food will Maria's pet have eaten than Trenton's pet will have eaten over 2 seven-day weeks

Step-by-step explanation:

We have , Trenton and Maria record how much dry food their pets eat on average each day.• Trenton's pet: 4/5 cup of dry food• Maria's pet: 1.25 cups of dry food. Based on these averages . We need to find how many more cups of dry food will Maria's pet have eaten than Trenton's pet will have eaten over 2 seven-day weeks . We need to find how much they eat for 14 days as:

Trenton's pet: 4/5 cup of dry food•

With 4/5 per day , for 14 days :

⇒ $14(\frac{4}{5} )$

⇒ $14(0.8 )$

⇒ $11.2$

Maria's pet: 1.25 cups of dry food.

With 1.25 per day , for 14 days :

⇒ $14(1.25 )$

⇒ $17.5$

Subtracting Maria's - Trenton's :

⇒ $17.5-11.2=6.3$

That means , 6.3 many more cups of dry food will Maria's pet have eaten than Trenton's pet will have eaten over 2 seven-day weeks

3 0

3 years ago

How many 1/16 pound servings of chocolate are in a 1/2 pound chocolate bar

Keith_Richards [23]

1/2 = 8/16 . . . so there are (8) - 1/16 pounds servings of chocolate in a 1/2 pound chocolate bar

6 0

3 years ago

What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean

Elena-2011 [213]

Answer:

0.658 is the probability that a sample 90 test takers will provide a sample mean test score within 10 points of the population mean of 502.

Step-by-step explanation:

The following information is missing:

The standard deviation of population is 100.

We are given the following information in the question:

Population mean, μ = 502

Standard Deviation, σ = 100

Sample size, n = 90

Standard error =

$\dfrac{\sigma}{\sqrt{n}} = \dfrac{100}{\sqrt{90}} = 10.54$

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(test score within 10 points)

$P(492 \leq x \leq 512) \\\\= P(\displaystyle\frac{492 - 502}{10.54} \leq z \leq \displaystyle\frac{512-502}{10.54}) \\\\= P(-0.9487 \leq z \leq 0.9487)\\= P(z \leq 0.9487) - P(z < -0.9487)\\= 0.829 -0.171 = 0.658 = 65.8\%$