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e-lub [12.9K]
3 years ago
12

Solve the four problems below. Which of the following problems can be solved by finding ? 3 divided by 1/2

Mathematics
1 answer:
Anastasy [175]3 years ago
7 0

Answer:

(a)=6\\(b)=1\ \frac {1}{2}\\(c)=1\ \frac {1}{2}\\d=6

Step-by-step explanation:

(a)

3 ft long when divided into half ft each will be

3ft ÷1/2 ft

We change the ÷ to × then the second part is reciprocated hence

3ft÷1/2= 3ft×2/1=6 pieces

(b)

The question requires us to get the differences. Initially, Phil makes 3 quarts of soup but the family takes only half of the original quantity to mean Phil remains with the other half.

Therefore, half of 3 is 3×1/2=3/2= 1.5 or 1 1/2

(c)

Hiding 3 pounds of gold into two equal sections means 3 is divided by 2

3÷2=1.5 or 1 1/2

(d)

If half, 0.5 represent 3 cups then full quantity represent x

0.5=3 cups

1=x

By cross multiplication

1*3/0.5=6

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The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of 10 batteries
Nat2105 [25]

Answer:

(1) We are conducting the one-sample t-test and will be testing for the hypothesis of mean battery greater than 25 h. So, we will reject the null hypothesis for mean battery life equal to 25 hr against the alternative hypothesis for mean battery life greater than 25 hr.

(2) A 95% lower confidence interval on mean battery life is 25.06 hr.

Step-by-step explanation:

We are given that a random sample of 10 batteries is subjected to an accelerated life test by running them continuously at an elevated temperature until failure, and the following lifetimes (in hours) are obtained: 25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

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

where, \bar X = sample mean BAC = \frac{\sum X}{n} = 26 hr

             s = sample standard deviation = \sqrt{\frac{\sum(X - \bar X)^{2} }{n-1} } = 1.625 hr

             n = sample of batteries = 10

             \mu = population mean battery life

<em> Here for constructing a 95% lower confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

(1) It is stated that the manufacturer wants to be certain that the mean battery life exceeds 25 h.

Since we are conducting the one-sample t-test and will be testing for the hypothesis of mean battery greater than 25 h. So, we will reject the null hypothesis for mean battery life equal to 25 hr against the alternative hypothesis for mean battery life greater than 25 hr.

(2) So, a 95% lower confidence interval for the population mean, \mu is;

P(-1.833 < t_9) = 0.95  {As the lower critical value of t at 9

                                             degrees of  freedom is -1.833 with P = 5%}    

P(-1.833 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }) = 0.95

P( -1.833 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} ) = 0.95

P( \bar X-1.833 \times {\frac{s}{\sqrt{n} } } < \mu) = 0.95

<u>95% lower confidence interval for</u> \mu = [ \bar X-1.833 \times {\frac{s}{\sqrt{n} } } ]

                                                             = [ 26-1.833 \times {\frac{1.625}{\sqrt{10} } } ]

                                                            = [25.06 hr]

Therefore, a 95% lower confidence interval on mean battery life is 25.06 hr.

5 0
3 years ago
Plz help me can you explain to me have to work this problems.
Amanda [17]
Is This Elementary School HomeWork Or High School HomeWork
8 0
4 years ago
Read 2 more answers
In six months, 33.6 gallons were used. Find the unit rate.
Nana76 [90]

I believe the answer is 5.6

Just divide 33.6 by 6

6 0
4 years ago
The heights of male are normally distributed with mean of 170 cm and standard deviation
Bumek [7]

Answer:

0.0918 = 9.18% probability that a randomly selected male has a height > 180 cm.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 170cm and standard deviation of 7.5 cm.

This means that \mu = 170, \sigma = 7.5

Find the probability that a randomly selected male has a height > 180 cm.

This is 1 subtracted by the pvalue of Z when X = 180. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{180 - 170}{7.5}

Z = 1.33

Z = 1.33 has a pvalue of 0.9082

1 - 0.9082 = 0.0918

0.0918 = 9.18% probability that a randomly selected male has a height > 180 cm.

4 0
3 years ago
Aaron has a digital scale. He puts a marshmallow on The digital scale and it reads 7.2 grams. How much would you except 10 marsh
madreJ [45]

Answer:

72 grams

Step-by-step explanation:

If one marshmallow weighs 7.2 all you have to do is take that times 10, which would give you 72 grams as the answer.

4 0
4 years ago
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