Let f(x) = -x-1, h(x) = x-3/3 Find (f o h)(-1).

1.5x+0.75y what does x variable

Illusion [34]

Answer:

1.5

Step-by-step explanation:

3 0

3 years ago

The soccer team can have no more than 26 players. How many more players can make the team if the coach has already chosen 12 pla

o-na [289]

Answer: >14 I think

Explanation:

3 0

3 years ago

What is the perimeter of this figure?

krok68 [10]

Answer:

D) 44 inches

Step-by-step explanation:

Isosceles trapezoids have equal legs, which is why they are called isosceles.

Squares have equal side lengths are around, so you know that the missing side lengths to this figure are 5 in, 10 in, and 10 inches.

Add all of the given values together.

5 in + 4 in + 10 in = 19

Add all of the "unknown" values together.

5 in + 10 in + 10 in = 25

Add these two values together.

19 in + 25 in = 44 inches

This answer corresponds to answer choice D.

4 0

2 years ago

Read 2 more answers

Triangle ABC has vertices A(-5, -2), B(7, -5), and C(3, 1). Find the coordinates of the intersection of the three altitudes

Darina [25.2K]

Answer:

Orthocentre (intersection of altitudes) is at (37/10, 19/5)

Step-by-step explanation:

Given three vertices of a triangle

A(-5, -2)

B(7, -5)

C(3, 1)

Solution A by geometry

Slope AB = (yb-ya) / (xb-xa) = (-5-(-2)) / (7-(-5)) = -3/12 = -1/4

Slope of line normal to AB, nab = -1/(-1/4) = 4

Altitude of AB = line through C normal to AB

(y-yc) = nab(x-xc)

y-1 = (4)(x-3)

y = 4x-11 .........................(1)

Slope BC = (yc-yb) / (xc-yb) = (1-(-5) / (3-7)= 6 / (-4) = -3/2

Slope of line normal to BC, nbc = -1 / (-3/2) = 2/3

Altitude of BC

(y-ya) = nbc(x-xa)

y-(-2) = (2/3)(x-(-5)

y = 2x/3 + 10/3 - 2

y = (2/3)(x+2) ........................(2)

Orthocentre is at the intersection of (1) & (2)

Equate right-hand sides

4x-11 = (2/3)(x+2)

Cross multiply and simplify

12x-33 = 2x+4

10x = 37

x = 37/10 ...................(3)

substitute (3) in (2)

y = (2/3)(37/10+2)

y=(2/3)(57/10)

y = 19/5 ......................(4)

Therefore the orthocentre is at (37/10, 19/5)

Alternative Solution B using vectors

Let the position vectors of the vertices represented by

a = <-5, -2>

b = <7, -5>

c = <3, 1>

and the position vector of the orthocentre, to be found

d = <x,y>

the line perpendicular to BC through A

(a-d).(b-c) = 0 "." is the dot product

expanding

<-5-x,-2-y>.<4,-6> = 0

simplifying

6y-4x-8 = 0 ...................(5)

Similarly, line perpendicular to CA through B

<b-d>.<c-a> = 0

<7-x,-5-y>.<8,3> = 0

Expand and simplify

-3y-8x+41 = 0 ..............(6)

Solve for x, (5) + 2(6)

-20x + 74 = 0

x = 37/10 .............(7)

Substitute (7) in (6)

-3y - 8(37/10) + 41 =0

3y = 114/10

y = 19/5 .............(8)

So orthocentre is at (37/10, 19/5) as in part A.

8 0

3 years ago

According to an NRF survey conducted by BIGresearch, the average family spends about $237 on electronics (computers, cell phones

Usimov [2.4K]

Answer:

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.

Step-by-step explanation:

We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.

Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.

Let X = back-to-college family spending on electronics

SO, X ~ Normal( $\mu=237,\sigma^{2} =54^{2}$ )

The z score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean family spending = $237

$\sigma$ = standard deviation = $54

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)

P(X < $150) = P( $\frac{X-\mu}{\sigma}$ < $\frac{150-237}{54}$ ) = P(Z < -1.61) = 1 - P(Z $\leq$ 1.61)

= 1 - 0.9463 = 0.0537

The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)

P(X > $390) = P( $\frac{X-\mu}{\sigma}$ > $\frac{390-237}{54}$ ) = P(Z > 2.83) = 1 - P(Z $\leq$ 2.83)

= 1 - 0.9977 = 0.0023

The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)

P($120 < X < $175) = P(X < $175) - P(X $\leq$ $120)

P(X < $175) = P( $\frac{X-\mu}{\sigma}$ < $\frac{175-237}{54}$ ) = P(Z < -1.15) = 1 - P(Z $\leq$ 1.15)

= 1 - 0.8749 = 0.1251

P(X < $120) = P( $\frac{X-\mu}{\sigma}$ < $\frac{120-237}{54}$ ) = P(Z < -2.17) = 1 - P(Z $\leq$ 2.17)

= 1 - 0.9850 = 0.015

The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.

Therefore, P($120 < X < $175) = 0.1251 - 0.015 = 0.1101

5 0

3 years ago