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3 years ago

Hi . I need help with this question. Please show workings.

Mathematics

2 answers:

dybincka [34]3 years ago

8 0

Answer:

6km/h

Step-by-step explanation:

$speed \: = \frac{distance}{time} \\ 50m \: = 0.5km \\ 30s \: = 0.0833333 \: hours$

$speed \: = \frac{0.5}{0.0833333} \\ speed \: = 6.0$

marishachu [46]3 years ago

5 0

Answer:

Step-by-step explanation:

Solution,

Distance travelled = 50 m

Time taken = 30 second

Let's find the speed in meter/ second

Speed = $\frac{distance \: travelled \: }{time \: taken}$

plug the values

$= \frac{50}{30}$

$= \frac{5}{3 \: } \: meter \: per \: second$

Now,Let's find the speed in km/ hour

$\frac{5}{3} \times \frac{3600}{1000}$

Reduce the numbers with G.C.F 3

$5 \times \frac{1200}{1000}$

Reduce the numbers with GCF 5

$\frac{1200}{200}$

Reduce the fraction with 200

$6 \: kilometer \: per \: hour$

Hope this helps..

Best regards!

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Because of staffing decisions, managers of the Gibson-Marion Hotel are interested in the variability in the number of rooms occu

olchik [2.2K]

Answer:

a) $s^2 =30^2 =900$

b) $\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}$

$567.28 \leq \sigma^2 \leq 1690.224$

c) $23.818 \leq \sigma \leq 41.112$

Step-by-step explanation:

Assuming the following question: Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in the variability in the number of rooms occupied per day during a particular season of the year. A sample of 20 days of operation shows a sample mean of 290 rooms occupied per day and a sample standard deviation of 30 rooms

Part a

For this case the best point of estimate for the population variance would be:

$s^2 =30^2 =900$

Part b

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom are given by:

$df=n-1=20-1=19$

Since the Confidence is 0.90 or 90%, the significance $\alpha=0.1$ and $\alpha/2 =0.05$ , the critical values for this case are:

$\chi^2_{\alpha/2}=30.144$

$\chi^2_{1- \alpha/2}=10.117$

And replacing into the formula for the interval we got:

$\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}$

$567.28 \leq \sigma^2 \leq 1690.224$

Part c

Now we just take square root on both sides of the interval and we got:

$23.818 \leq \sigma \leq 41.112$

5 0

3 years ago

Select the linear equation represented in the graph

LiRa [457]

Answer:

Step-by-step explanation:

Since the line in the picture rises 1 unit for every 2 units it moves to the right, it has a slope of 1/2. Also, since it intersects the y axis at (0,3), the y intercept is 3. Therefore, the equation of the line is y=1/2x+3, or answer choice B. Hope this helps!

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2 years ago

Prove algebraically that 0.5 recurring = 5/9

weeeeeb [17]

Answer:

see explanation

Step-by-step explanation:

We require 2 equations with the recurring part after the decimal point.

Let x = 0.555..... → (1)

Multiply both sides by 10, then

10x = 5.555..... → (2)

Subtract (1) from (2) thus eliminating the recurring decimal

(10x - x) = (5.5555 - 0.5555 ), that is

9x = 5 ( divide both sides by 9 )

x = $\frac{5}{9}$

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2 years ago

The two sets of lines are parallel. If the measure of angle A is 110°, what is the measure of angle B?

Mrrafil [7]

Answer:

Sum of 2 interior angles =180

1 angle+70=180

Another angle=110

Step-by-step explanation:

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2 years ago

Which of the following expenses is not a variable cost?

Sindrei [870]

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2 years ago

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Hi . I need help with this question. Please show workings.​

Hi . I need help with this question. Please show workings.