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sergeinik [125]

2 years ago

11

The bottom of Ignacio's desktop is 74.5\,\text{cm}74.5cm74, point, 5, space, c, m from the floor. Ignacio sits in his adjustable

chair, and the tops of his legs are 49.3\,\text{cm}49.3cm49, point, 3, space, c, m from the floor. Each clockwise rotation of the knob on the chair raises Ignacio's legs by 4.8\,\text{cm}4.8cm4, point, 8, space, c, m.
Write an inequality to determine the number of clockwise rotations, rrr, Ignacio could make with the knob without his legs touching the desk.

1 answer:

Musya8 [376]2 years ago

7 0

Answer:

The inequality is:

49.3+4.8r <74.5

The solution set is r < 5.25

Step-by-step explanation:

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Two families went to Rollercoaster World. The Brown family paid 170 for 3 children and 2 adults. The Peckham family paid 360 for

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Answer: the system of equations are

3x + 2y = 170

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Step-by-step explanation:

Let x represent the price of a child's ticket in dollars.

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2 years ago

The national average for mathematics SATs in 2011 was 514 and the standard deviation was approximately 40. a) Within what bounda

blondinia [14]

Answer:

$0.75 = 1-\frac{1}{k^2}$

If we solve for k we can do this:

$\frac{1}{k^2}= 1-0.75=0.25$

$\frac{1}{0.25}= k^2$

$k^2 =4$

$k =\pm 2$

So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

$Lower = \mu -2\sigma = 514- 2*40=434$

$Upper = \mu +2\sigma = 514 + 2*40=594$

Step-by-step explanation:

We don't know the distribution for the scores. But we know the following properties:

$\mu = 514 , \sigma =40$

For this case we can use the Chebysev theorem who states that "At least $1 -\frac{1}{k^2}$ of the values lies between $\mu -k\sigma$ and $\mu +k\sigma$ "

And we need the boundaries on which we expect at least 75% of the scores. If we use the Chebysev rule we have this:

$0.75 = 1-\frac{1}{k^2}$

If we solve for k we can do this:

$\frac{1}{k^2}= 1-0.75=0.25$

$\frac{1}{0.25}= k^2$

$k^2 =4$

$k =\pm 2$

So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

$Lower = \mu -2\sigma = 514- 2*40=434$

$Upper = \mu +2\sigma = 514 + 2*40=594$

4 0

2 years ago

Line segment JL is an altitude in triangle JKM. Which statement explains whether JKM is a right triangle? Round measures to the

natulia [17]

Answer:

C. JKM is not a right triangle because KM ≠ 15.3.

Step-by-step explanation:

We can see from our diagram that triangle JKM is divided into right triangles JLM and JLK.

In order to triangle JKM be a right triangle $KM^{2}=JK^{2}+JM^{2}$ .

We will find length of side KM using our right triangles JLM and JLK as $KM=KL+LM$ .

Using Pythagorean theorem in triangle JLM we will get,

$LM=\sqrt{JM^{2}-JL^{2}}$

$LM=\sqrt{8^{2}-5^{2}}$

$LM=\sqrt{64-25}$

$LM=\sqrt{39}=6.244997998\approx 6.24$

Now let us find length of side KL.

$KL=\sqrt{JK^{2}-JL^{2}}$

$KL=\sqrt{13^{2}-5^{2}}$

$KL=\sqrt{169-25}$

$KL=\sqrt{144}=12$

Now let us find length of KM by adding lengths of KL and LM.

$KM=12+6.24=18.24$

Now let us find whether JKM is right triangle or not using Pythagorean theorem.

$KM^{2}=JK^{2}+JM^{2}$

$18.24^{2}=13^{2}+8^{2}$

$18.24^{2}=169+64$

$18.24^{2}=233$

Upon taking square root of both sides of equation we will get,

$18.24\neq 15.264337522473748$

$18.2\neq 15.3$

We have seen that KM equals 18.2 and in order to JKM be a right triangle KM must be equal to 15.3, therefore, JKM is not a right triangle and option C is the correct choice.

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2 years ago

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