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Olin [163]
3 years ago
6

Jason played golf at a rate of 6 holes in 24 minutes. How much longer would it take him to play 54 holes than 41 holes?

Mathematics
1 answer:
Mkey [24]3 years ago
5 0
6 holes in 24 mins.
1 hole in 4 mins.
54 × 4= 216
41 × 4= 164
216-164= 52
So it would take 52 more minutes.
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A 50 foot ladder is placed at 30 feet away from the building in order to reach the 3rd floor of the building. How high is the 3r
Sonbull [250]
The correct answer should be 20 feet
5 0
3 years ago
Somebody please help .. What is the exact volume of this right cone ?
stiks02 [169]
\bf \textit{volume of a cone}\\\\
V=\cfrac{\pi r^2 h}{3}~~
\begin{cases}
r=radius\\
h=height\\
-----\\
r=12\\
h=8
\end{cases}\implies V=\cfrac{\pi (12)^2(8)}{3}
7 0
3 years ago
The total budget for a landscaping project was $349.02, and the project called for 42 plants. What is the average cost per plant
Marianna [84]

Answer:

349.02 / 42 = answer

Step-by-step explanation:

method used is dividing total cost by number of goods used gives the exact cost of each plant

8 0
3 years ago
A manufacturer of chocolate candies uses machines to package candies as they move along a filling line. Although the packages ar
Elenna [48]

Answer:

We conclude that the mean amount packaged is equal to 8.17 ounces.

Step-by-step explanation:

We are given that in a particular sample of 50 packages, the mean amount dispensed is 8.171 ​ounces, with a sample standard deviation of 0.052 ounces.

Let \mu = <u><em>population mean amount packaged. </em></u>

So, Null Hypothesis, H_0 : \mu = 8.17 ounces    {means that the mean amount packaged is equal to 8.17 ounces}

Alternate Hypothesis, H_A : \mu\neq 8.17 ounces    {means that the mean amount packaged is different from 8.17 ounces}

The test statistics that will be used here is <u>One-sample t-test statistics</u> because we don't know about the population standard deviation;

                               T.S.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~  t_n_-_1

where, \bar X = sample mean amount dispensed = 8.171 ounces

             s = sample standard deviation = 0.052 ounces

            n = sample of packages = 50

So, <u><em>the test statistics</em></u> =  \frac{8.171-8.17}{\frac{0.052}{\sqrt{50} } }  ~   t_4_9

                                    =  0.1359  

The value of t-test statistics is 0.1359.

<u>Also, the P-value of test-statistics is given by;</u>

the meaning of the​ p-value is that the p-value is the probability of obtaining a sample mean that is equal to or more extreme than 0.001 ounces away from8.17 if the null hypothesis is true.

                    P-value = P( t_4_9 > 0.136) = More than 40% {from the t-table}

Since the P-value of our test statistics is more than the level of significance of 0.01, so <u><em>we have insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the mean amount packaged is equal to 8.17 ounces.

6 0
3 years ago
Are the two lines parallel, perpendicular, coinciding lines, or neither?
Elena L [17]

Answer:

D

Step-by-step explanation:

crosses at a 90 degree angle

6 0
3 years ago
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