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olya-2409 [2.1K]

3 years ago

13

In a certain computer company, 3/5 of the programmers know fortran, 1/6 know cobol, and 1/10 know both. what fraction know neith

er fortran nor cobol?

1 answer:

koban [17]3 years ago

6 0

Given that 3/5 know fortran
1/6 know cobol
1/10 know both fotran and cobol
the fraction of programmers who knows at least one of the two programs will be:
3/5+1/6+1/10
=13/15
the fraction of programmers who don't know either of the two will be:
1-13/15
=2/15
thus the answer is 2/15

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The manager of a local nightclub has recently surveyed a random sample of 280 customers of the club. She would now like to deter

Lunna [17]

Answer:

$z = \frac{35.6-35}{\frac{5}{\sqrt{280}}}= 2.007$

$p_v = P(z>2.007) = 0.0224$

Since the p value is lower than the significance level given of 0.05 we have enough evidence to reject the null hypothesis on this case. And the best conclusion for this case is:

We (reject) the null hypothesis. That means that we (found) evidence to support the alternative.

Step-by-step explanation:

We have the following info given:

$\bar X = 35.6$ represent the sampel mean for the age of customers

$\sigma = 5$ represent the population standard deviation

$n = 280$ represent the sample size selected

We want to test if the mean age of her customers is over 35 so then the system of hypothesis for this case are:

Null hypothesis: $\mu \leq 35$

Alternative hypothesis $\mu >35$

The statistic for this case is given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing the data given we got:

$z = \frac{35.6-35}{\frac{5}{\sqrt{280}}}= 2.007$

We can calculate the p value since we are conducting a right tailed test like this:

$p_v = P(z>2.007) = 0.0224$

Since the p value is lower than the significance level given of 0.05 we have enough evidence to reject the null hypothesis on this case. And the best conclusion for this case is:

We (reject) the null hypothesis. That means that we (found) evidence to support the alternative.

5 0

3 years ago

A garden is shaped in the form of a regular heptagon (seven-sided), MNSRQPO. A circle with center T and radius 25m circumscribes

Alenkinab [10]

The relationship between the sides MN, MS, and MQ in the given regular heptagon is $\dfrac{1}{MN} = \dfrac{1}{MS} + \dfrac{1}{MQ}$

The area to be planted with flowers is approximately <u>923.558 m²</u>

The reason the above value is correct is as follows;

The known parameters of the garden are;

The radius of the circle that circumscribes the heptagon, r = 25 m

The area left for the children playground = ΔMSQ

Required;

The area of the garden planted with flowers

Solution:

The area of an heptagon, is;

$A = \dfrac{7}{4} \cdot a^2 \cdot cot \left (\dfrac{180 ^{\circ}}{7} \right )$

The interior angle of an heptagon = 128.571°

The length of a side, S, is given as follows;

$\dfrac{s}{sin(180 - 128.571)} = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)}$

$s = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)} \times sin(180 - 128.571) \approx 21.69$

$The \ apothem \ a = 25 \times sin \left ( \dfrac{128.571}{2} \right) \approx 22.52$

The area of the heptagon MNSRQPO is therefore;

$A = \dfrac{7}{4} \times 22.52^2 \times cot \left (\dfrac{180 ^{\circ}}{7} \right ) \approx 1,842.94$

$MS = \sqrt{(21.69^2 + 21.69^2 - 2 \times 21.69 \times21.69\times cos(128.571^{\circ})) \approx 43.08$

By sine rule, we have

$\dfrac{21.69}{sin(\angle NSM)} = \dfrac{43.08}{sin(128.571 ^{\circ})}$

$sin(\angle NSM) =\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ})$

$\angle NSM = arcsin \left(\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ}) \right) \approx 23.18^{\circ}$

∠MSQ = 128.571 - 2*23.18 = 82.211

The area of triangle, MSQ, is given as follows;

$Area \ of \Delta MSQ = \dfrac{1}{2} \times 43.08^2 \times sin(82.211^{\circ}) \approx 919.382^{\circ}$

The area of the of the garden plated with flowers, $A_{req}$ , is given as follows;

$A_{req}$ = Area of heptagon MNSRQPO - Area of triangle ΔMSQ

Therefore;

$A_{req}$ = 1,842.94 - 919.382 ≈ 923.558

The area of the of the garden plated with flowers, $A_{req}$ ≈ <u>923.558 m²</u>

Learn more about figures circumscribed by a circle here:

brainly.com/question/16478185

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3 years ago

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Nutka1998 [239]

Answer:

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