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Step2247 [10]
3 years ago
11

What is the least common multiple of 5 and 150

Mathematics
2 answers:
kenny6666 [7]3 years ago
5 0

lcm = 150, since 5|150

lesya [120]3 years ago
3 0

Answer:

150

Step-by-step explanation:

5 x 25 = 125                150 x 1 =150

5 x 26= 130

5 x 27 = 135

5 x 28 = 140

5 x 29 = 145

5 x 30 = 150

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Aperson invest 10,000 shillings for 40 months compounded monthly. Find the accumulated amount If the rate is 5% per annum.​
yanalaym [24]

hi

monthly interest rate is :     \sqrt[12]{1.05}

So in forty month there will be  :  10 000 * ( \sqrt[12]{1.05} )^40  ≈11 776,06

4 0
1 year ago
Amy is planning the seating arrangement for her wedding reception. Each round table can sit 12 guests. The head table can sit th
storchak [24]

Answer:

23 tables.

Step-by-step explanation:

282- 8 = 274     274 divided by 12 = 22.8  round to the nearest whole number which is 23.

7 0
3 years ago
100 members of a Gym are surveyed. The results are shown below. Given the gym member chosen at random is a female, what is the p
Lina20 [59]

Answer:

\frac{17}{28}

Step-by-step explanation:

There are 56 females

There is 56 females

34 are 20+

34/56 simplfies to 17/28

8 0
3 years ago
A college infirmary conducted an experiment to determine the degree of relief provided by three cough remedies. Each cough remed
KiRa [710]

Answer:

p_v = P(\chi^2_{4,0.05} >3.81)=0.43233

Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed. So we can say that the 3 remedies ar approximately equally effective.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

                               NyQuil          Robitussin       Triaminic     Total

No relief                    11                     13                      9               33

Some relief               32                   28                     27              87

Total relief                7                       9                      14               30

Total                         50                    50                    50              150

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference in the three remedies

H1: There is a difference in the three remedies

The level os significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{total col * total row}{grand total}

And the calculations are given by:

E_{1} =\frac{50*33}{150}=11

E_{2} =\frac{50*33}{150}=11

E_{3} =\frac{50*33}{150}=11

E_{4} =\frac{50*87}{150}=29

E_{5} =\frac{50*87}{150}=29

E_{6} =\frac{50*87}{150}=29

E_{7} =\frac{50*30}{150}=10

E_{8} =\frac{50*30}{150}=10

E_{9} =\frac{50*30}{150}=10

And the expected values are given by:

                               NyQuil          Robitussin       Triaminic     Total

No relief                    11                     11                       11               33

Some relief               29                   29                     29              87

Total relief                10                     10                     10               30

Total                         50                    50                    50              150

And now we can calculate the statistic:

\chi^2 = \frac{(11-11)^2}{11}+\frac{(13-11)^2}{11}+\frac{(9-11)^2}{11}+\frac{(32-29)^2}{29}+\frac{(28-29)^2}{29}+\frac{(27-29)^2}{29}+\frac{(7-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(14-10)^2}{10} =3.81

Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(3-1)(3-1)=4

And we can calculate the p value given by:

p_v = P(\chi^2_{4,0.05} >3.81)=0.43233

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(3.81,4,TRUE)"

Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed.

4 0
3 years ago
Noelle ran 4 1/2 miles in the morning and 3 1/4 miles in the evening. How far did Noelle run that day?
KIM [24]
You just add all the numbers together:
3 1/8+ 4 1/2+ 4 1/4
=25/8+9/2+17/4
=25/8+36/8+34/8
=95/8
=11 7/8

Ugh.. did you type in the numbers correctly? My guess would be it was meant to be B tho :)

6 0
3 years ago
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