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2 years ago

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Which of the following is the graph of y = StartRoot negative x minus 3 EndRoot? On a coordinate plane, an absolute value curve

opens up and to the right in quadrant 4. It starts at x = 3. On a coordinate plane, an absolute value curve opens down and to the left in quadrants 1 and 2. It starts at x = 3. On a coordinate plane, an absolute value curve opens down and to the left in quadrant 2. It starts at x = negative 3. On a coordinate plane, an absolute value curve opens up and to the right in quadrants 3 and 4. It starts at x = negative 3.

1 answer:

MariettaO [177]2 years ago

8 0

Answer:

d

Step-by-step explanation:

just took the test

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Ted creates a box plot using 14, 13, 21, 10, 28, 30, and 35 as the data. Which of the following box plots shows the data accurat

Slav-nsk [51]

<span> A box plot is drawn with end points at 10 and 35.The box extends from 13 to 30 and a vertical line is drawn inside the box at 21.</span>

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2 years ago

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Use the drop-down menus to complete the proof. Given that w ∥ x and y is a transversal, we know that ∠1 ≅∠5 by the . Therefore,

ololo11 [35]

Answer:

From the graph attached, we know that $\angle 1 \cong \angle 5$ by the corresponding angle theorem, this theorem is about all angles that derive form the intersection of one transversal line with a pair of parallels. Specifically, corresponding angles are those which are placed at the same side of the transversal, one interior to parallels, one exterior to parallels, like $\angle 1$ and $\angle 5$ .

We also know that, by definition of linear pair postulate, $\angle 3$ and $\angle 1$ are linear pair. Linear pair postulate is a math concept that defines two angles that are adjacent and for a straight angle, which is equal to 180°.

They are supplementary by the definition of supplementary angles. This definition states that angles which sum 180° are supplementary, and we found that $\angle 3$ and $\angle 1$ together are 180°, because they are on a straight angle. That is, $m \angle 3 + m \angle 1 = 180\°$

If we substitute $\angle 5$ for $\angle 1$ , we have $m \angle 3 + m \angle 5 = 180\°$ , which means that $\angle 3$ and $\angle 5$ are also supplementary by definition.

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2 years ago

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A 2-column table with 6 rows. The first column is labeled miles driven with entries 27, 65, 83, 109, 142, 175. The second column

Helen [10]

Answer:

1) The linear regression model is y = -0.0348·x + 13.989

2) The correlation coefficient is -0.0725

3) The strength of the model is strong - association

Step-by-step explanation:

1)

X Y XY X²

27 13 351 729

65 12 780 4225

83 11 913 6889

109 10 1090 11881

142 9 1278 20164

175 8 1400 30625

∑ 601 63 5812 74513

From y = ax + b, we have

$a = \frac{n\sum xy - \sum x\sum y }{n\sum x^{2}-\left (\sum x \right )^{2}} = \frac{6 \times 5812 - 601 \times 63}{6 \times 74513-601^{2}} = - 0.0348$

b = 1/n(∑y -a∑x) = 1/6(63 - (0.0348)×601) = 13.989

Therefore, the linear regression model is y = -0.0348·x + 13.989

2)

$r = \frac{n\sum xy - \sum x\sum y }{\sqrt{[n\sum x^{2}-\left (\sum x \right )^{2}] [n\sum y^{2}-\left (\sum y \right )^{2}]}} = \frac{6 \times 5812 - 601 \times 63}{\sqrt{[6 \times 74513-601^{2}] [6 \times 3969 - 63^2]} } = - 0.0725$

3) The strength is - association.

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2 years ago

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Out of 498 applicants for a job, 187 have over 10 years of experience and 127 have over 10 years of experience and have a gradua

Kitty [74]

Answer:

the probability that randomly selected applicants over 10 years of experience is 0.6791

Step-by-step explanation:

The computation of the probability that randomly selected applicants over 10 years of experience is as follows:

Total would be

= 187 have 10 + years experience

Now

P(graduate | 10+) = (graduate and 10+ years experience) ÷ (10 + years of experience)

= 127 ÷ 187

= 0.6791

Hence, the probability that randomly selected applicants over 10 years of experience is 0.6791

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1 year ago

Player V and Player M have competed against each other many times. Historical data show that each player is equally likely to wi

Helen [10]

Answer:

0.46 (46%)

Step-by-step explanation:

We have the following data:

- Probability that player V wins the first set:

$p(1)=0.50$

(because the text says the two players are equally likely to win the first set)

- Probability that player V wins the 2nd set if he has won the 1st set:

$p(2|1)=0.60$

So, the probability that player V wins the first 2 sets is:

$p(12)=p(1)\cdot p(2|1)=(0.50)(0.60)=0.30$ (1)

Instead, the probability that player V loses the 2nd set if he has won the 1st set is 0.40 (=1-0.60), so

$p(2^c|1)=0.40$

So, the probabiity that player V winse the 1st set but loses the 2nd set is

$p(12^c)=p(1)\cdot p(2^c|1)=(0.50)(0.40)=0.20$ (2)

Also, we have:

- Probability that player V loses the 1st set:

$p(1^c)=0.50$

- Probability that she will lose the 2nd set in this case is 0.70, it means that the probability that she will win the 2nd set if she lost the 1st set is 0.30, so:

$p(2|1^c)=0.30$

So, the probability that she will lose the 1st set and win the 2nd set is:

$p(1^c2)=p(1^c)\cdot p(2|1^c)=(0.50)(0.30)=0.15$ (3)

Combined together (2) and (3), this means that the probability that player V wins exactly 1 set out of the first two sets is:

$p(1/2)=p(12^c)+p(1^c2)=0.20+0.15=0.35$ (4)

At this point, the probability that she will win the 3rd set is

$p(3)=0.45$

This means that the overall probability that she will win the 3rd set if she won exacty 1 of the first 2 sets is:

$p(1/2,3) = p(1/2)\cdot p(3)=(0.35)(0.45)=0.16$ (5)

So, the overall probability that player V will win a match against player M is the sum of (1) and (5):

$p(W)=p(12)+p(1/2,3)=0.30+0.16=0.46$

5 0

2 years ago