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3 years ago

11

An arts academy requires there to be 6 teachers for every 108 students and 3 tutors for every 36 students. students How many stu

dents does the academy have per teacher? Per tutor? How many tutors does the academy need if it has 84 students?

1 answer:

Alika [10]3 years ago

7 0

18 students = 1 teacher

13 students = 1 tutor

The academy needs 7 tutors for 84 students.

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To make 100g of jam you need 23g of strawberries, 51g of blackberries and the rest should be

Gelneren [198K]

More explain please?????

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3 years ago

Secant jkl and jmn are drawn to circle o from an external point ,j. if jk=8,lk=4 and jm=6 what is the length of jn answer

Archy [21]

See the picture attached to better understand the problem

we know that
If two secant segments are drawn to a <span>circle </span><span>from an exterior point, then the product of the measures of one secant segment and its external secant segment is equal to the product of the measures of the other secant segment and its external secant segment.
</span>so
jl*jk=jn*jm------> jn=jl*jk/jm

we have
<span>jk=8,lk=4 and jm=6
</span>jl=8+4----> 12

jn=jl*jk/jm-----> jn=12*8/6----> jn=16

the answer is
jn=16

6 0

3 years ago

The following probability distributions of job satisfaction scores for a sample of information systems (IS) senior executives an

Mashcka [7]

Answer:

Step-by-step explanation:

To calculate ;

1) the expected value of the job satisfaction score for senior executives ;

expected value = Summation (Px)

= 1 x 0.05 + 2 x 0.09 + 3 x 0.03 + 4 x 0.42 + 5 x 0.41

= 4.05

2) the expected value of the job satisfaction score for middle managers;

= 1 x 0.04 + 2 x 0.10 + 3 x 0.12 + 4 x 0.46 + 5 x 0.28

= 3.84

c) the variance of job satisfaction scores for executives and middle managers (to 2 decimals).

Executives ; Variance = Summation(PX^2 - Summation(PX)^2

i) For Executive Managers = 1 x 0.05 + 2^2 x 0.09 + 3^2 x 0.03 + 4^2 x 0.42 + 5^2 x 0.41 - 4.05^2 = 1.246 = 1.25

ii) for middle managers ; 1 x 0.04 + 2^2 x 0.10 + 3^2 x 0.12 + 4^2 x 0.46 + 5^2 x 0.28 - 3.84^2 = 1.134 = 1.13

d) the standard deviation of job satisfaction scores for both probability distributions (to 2 decimals). Executives, Middle managers;

For Executives = square root [ 1 x 0.05 + 2^2 x 0.09 + 3^2 x 0.03 + 4^2 x 0.42 + 5^2 x 0.41 - 4.05^2] = 1.12

For Middle Managers ; Square root [1 x 0.04 + 2^2 x 0.10 + 3^2 x 0.12 + 4^2 x 0.46 + 5^2 x 0.28 - 3.84^2 ] = 1.06

e) from the values gotten for the variance of both executive and middle managers, the variance of the former is more than that of the latter as such higher satisfaction with the executive managers.

5 0

3 years ago

Round to the nearest hundred thousand 257,098

olga nikolaevna [1]

300,000 because if a number is above 5, 50, 500, 5000, etc. then it rounds up instead of down.

4 0

3 years ago

Suppose the standard deviation of a normal population is known to be 3, and H0 asserts that the mean is equal to 12. A random sa

blagie [28]

Answer:

$z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9$

$p_v =P(z>1.9)=0.0287$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=12.95$ represent the sample mean

$\sigma=3$ represent the population standard deviation for the sample

$n=36$ sample size

$\mu_o =12$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 12, the system of hypothesis would be:

Null hypothesis: $\mu \leq 12$

Alternative hypothesis: $\mu > 12$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9$

P-value

Since is a right tailed test the p value would be:

$p_v =P(z>1.9)=0.0287$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.

4 0

3 years ago

Read 2 more answers