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3 years ago

14

hugo sells handmade key chains online. Yesterday he made 6 keychains in 30minutes. hugo wants to make 9 keychains today . if hug

o works at the same rate today how long will it take him to make the keychains

2 answers:

Art [367]3 years ago

8 0

Answer:

45 minutes, hope this helps

Step-by-step explanation:

8090 [49]3 years ago

8 0

Answer:

45 minutes.

Step-by-step explanation:

Set up a ratio.

6 keychains: 30 minutes

divide both sides of the ratio by 6

1 keychain: 5 minutes

multiply both sides of the ratio by 9

9 keychains: 45 minutes

It will take him 45 minutes.

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Last year Betty's annual salary was $42000. This year she received a raise and now eams $46500. On a weekly basis, how much more

bearhunter [10]

Answer:

About 87$

Step-by-step explanation:

Yu take both salary amounts an divide by 52 since there are 52 weeks in a year and then find the difference between them

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3 years ago

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Rama09 [41]

Answer:

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Step-by-step explanation:

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3 years ago

WILL GIVE BRAINLY!!!! describe the types of correlation between two variables on a graph

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3 years ago

Read 2 more answers

Listed below are the number of years it took for a random sample of college students to earn bachelor's degrees (based on data f

schepotkina [342]

Answer:

a) Mean = 6.5, sample standard deviation = 3.50

b) Standard error = 0.7826

c) Point estimate = 6.5

d) Confidence interval: (5.1469 ,7.8531)

Step-by-step explanation:

We are given the following data set for students to earn bachelor's degrees.

4, 4, 4, 4, 4, 4, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 6, 6, 8, 9, 9, 13, 13, 15

a) Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{130}{20} = 6.5$

Sum of squares of differences = 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 4 + 4 + 4 + 4 + 4 + 4 + 0.25 + 0.25 + 2.25 + 6.25 + 6.25 + 42.25 + 42.25 + 72.25 = 233.5

$S.D = \sqrt{\frac{233.5}{19}} = 3.50$

b) Standard Error

$= \displaystyle\frac{s}{\sqrt{n}} = \frac{3.50}{\sqrt{20}} = 0.7826$

c) Point estimate for the mean time required for all college is given by the sample mean.

$\bar{x} = 6.5$

d) 90% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.10} = \pm 1.729$

$6.5 \pm 1.729(\frac{3.50}{\sqrt{20}} ) = 6.5 \pm 1.3531 = (5.1469 ,7.8531)$

e) No, the confidence interval does not contain the value of 4 years. Thus, confidence interval is not a good estimator as most of the value in the sample is of 4 years. Most of the sample does not lie in the given confidence interval.

5 0

4 years ago

Sabiendo que cada persona unos 4 500 000 glóbulos rojos por milímetro cúbico y que su cantidad de sangre es de 5 litros A) Que l

love history [14]

Answer:

Part A) distance covered by the aligned red cells is: 180000 Km

Part B) 8000000 bacteria (8 million bacteria)

Step-by-step explanation:

Part A)

Let's first find the total number of red cells that the person has in the 5 liters, knowing that there are 4,500,000 of them in a cubic millimetre:

Since 1 liter contains 1,000,000 cubic millimetres, then, in 5 liters there must be 5 times that number of cubic millimetres. That is a total of 5,000,000 cubic millimetres.

Since in each cubic millimetre there are 4,500,000 red cells, then the number of red cells in the 5 liters must be the product of these two quantities:

4,500,000 x 5,000,000 = 22,500,000,000,000 red blood cells.

Now, to find the distance they will cover aligned one after the other, knowing that their average diameter is 0.008 mm is:

22,500,000,000,000 x 0.008 mm= 180000000000 mm

$22.5\,\,10^{12} * 8\,\,10^{-3}=1.8\,\,10^{11}$ mm

In order to convert this distance into Km as requested, we use the fact that 1 mm = 0.000001 Km, or using exponents in base ten:

1 mm = $10^{-6}\,\,Km$

Then, the distance covered in kilometers is:

$1.8\,\,10^{11}\,*\,10^{-6} \,\,Km=1.8\,\,10^{5}\,\,Km=180000\,\,Km$

Part B)

The vaccine contains a total of $10^8$ bacteria per $cm^3$

In order to find how many bacteria we have in 80 $mm^3$ of the vaccine, we first need to convert 80 $mm^3$ into $cm^3$ :

$80\,\, mm^3 = 0.08 \,\,cm^3$

Then, if in one $cm^3$ there is $10^8$ bacteria, in $0.08\,\,cm^3$ there would be: $0.08\,*\,10^8$ bacteria. That is: $8\,\,10^6$ bacteria = 8000000 bacteria (eight million bacteria)

8 0

3 years ago