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2 years ago

In a school of 1600 students, the ratio of teachers to students is 1:16.

Mathematics

1 answer:

a_sh-v [17]2 years ago

4 0

Answer:

The number of teacher who joined the school is 20

Step-by-step explanation:

Teacher : Students = 1 : 16

Number of teachers = 1x

Number of students = 16x

16x = 1600

x = 1600/16

x = 100

Number of teachers at the beginning = 100

Now, after some teachers (let it be y) joining the school,

Teacher : students = 3:40

$\frac{100+y}{1600}=\frac{3}{40}\\\\40*(100+y)=3*1600\\\\4000+40y = 4800\\\\40y=4800-4000\\\\40y=800\\\\y=\frac{800}{40}\\\\y=20$

The number of teacher who joined the school is 20

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2 years ago

If DE = 4x + 10, EF = 2x-1, and DF = 9x-15, find DF

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Hello,
We have that:
(4x+10)+(2x-1)=9x-15

We solve the equation:
4x+10+2x-1=9x-15;
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The hourly wage increases each employee receives each year depends on the number of years of service. Every three years of servi

11Alexandr11 [23.1K]

Answer:

________{0.50 if x < 3

________{1.00 if 3 ≤ x < 6

f(x) = ____{1.50 if 6 ≤ x < 9

________{2.00 if 9 ≤ x < 12

Step-by-step explanation:

Given the information above :

Less than Employees = $0.5 increase per hour

Atleast 3 but less than 6 years employees = $1.00 increase per hour

Atleast 6 but less than 9 years employees = $1.50 increase per hour

Atleast 9 but less than 12 years employees = $2.00 increase per hour

The information above can be written as a piecewise function :

________{0.50 if x < 3

________{1.00 if 3 ≤ x < 6

f(x) = ____{1.50 if 6 ≤ x < 9

________{2.00 if 9 ≤ x < 12

The constraints is represented in the piecewise function above with x being the number of years since employee has been in service.

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1 year ago

Last year, 46% of business owners gave a holiday gift to their employees. A survey of business owners indicated that 35% plan to

DiKsa [7]

Answer:

a) Number = 60 *0.35=21

b) Since is a left tailed test the p value would be:

$p_v =P(Z$

c) If we compare the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% the proportion of business owners providing holiday gifts had decreased from the 2008 level.

We can use as smallest significance level 0.044 and we got the same conclusion.

Step-by-step explanation:

Data given and notation

n=60 represent the random sample taken

X represent the business owners plan to provide a holiday gift to their employees

$\hat p=0.35$ estimated proportion of business owners plan to provide a holiday gift to their employees

$p_o=0.46$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

Part a

On this case w ejust need to multiply the value of th sample size by the proportion given like this:

Number = 60 *0.35=21

2) Part b

We need to conduct a hypothesis in order to test the claim that the proportion of business owners providing holiday gifts had decreased from the 2008 level:

Null hypothesis: $p\geq 0.46$

Alternative hypothesis: $p < 0.46$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.35 -0.46}{\sqrt{\frac{0.46(1-0.46)}{60}}}=-1.710$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v =P(Z$

Part c

If we compare the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% the proportion of business owners providing holiday gifts had decreased from the 2008 level.

We can use as smallest significance level 0.044 and we got the same conclusion.

8 0

2 years ago