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3 years ago

Find the axis of symmetry, vertex, domain, and range of the following quadratic function. (show work pls)

Mathematics

1 answer:

ratelena [41]3 years ago

4 0

Answer:

All I know is the axis is horizontal lol

Step-by-step explanation:

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Find x and y and i just need the answer, no need to show work.

vichka [17]

Answer:

x= 20

y = 60

Step-by-step explanation:

y = 3x

6(3x) = 360

3x = 360/6

3x = 60

x = 20

y = 3*20

y = 60

7 0

1 year ago

Point K lies at (-3,4). Point K is reflected over the y-axis and then translated 7 units down to produce K’. Which rule could ha

bogdanovich [222]

Answer:

K(x, y) —> K’(-x, y - 7)

Step-by-step explanation:

7 0

2 years ago

there are 103 students eating lunch in the cafeteria. each table seats 4 students. all the tables are full except for 1 table. h

natta225 [31]

4*25= 100. There are 26 tables. At each of the 25 tables 4 people are sitting. Which means 100 people are sitting in complete tables of 4 people. That means that at the 26th table 3 people are sitting. 103-100= 3. Hope that helps. :)

6 0

3 years ago

Read 2 more answers

2/5 divided by 8/15 in its simplest form

gregori [183]

Step-by-step explanation:

2/5 ÷ 8/15

2/5 × 15/8

=3/4

Hope it helps ya

5 0

2 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Anna [14]

Answer:

a) 0.4452

b) 0.0548

c) 0.0501

d) 0.9145

e) 6.08 minutes or greater

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.7 minutes

Standard Deviation, σ = 0.50 minutes.

We are given that the distribution of length of the calls is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(calls last between 4.7 and 5.5 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{4.7 - 4.7}{0.50} \leq z \leq \displaystyle\frac{5.5-4.7}{0.50}) = P(0 \leq z \leq 1.6)\\\\= P(z \leq 1.6) - P(z$

$P(4.7 \leq x \leq 5.5) = 44.52\%$

b) P(calls last more than 5.5 minutes)

$P(x > 5.5) = P(z > \displaystyle\frac{5.5-4.7}{0.50}) = P(z > 1.6)\\\\P( z > 1.6) = 1 - P(z \leq 1.6)$

Calculating the value from the standard normal table we have,

$1 - 0.9452 = 0.0548 = 5.48\%\\P( x > 5.5) = 5.48\%$

c) P( calls last between 5.5 and 6 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{5.5 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(1.6 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$

$P(5.5 \leq x \leq 6) = 5.01\%$

d) P( calls last between 4 and 6 minutes)

$P(4 \leq x \leq 6) = P(\displaystyle\frac{4 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(-1.4 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$