Select the ordered pairs that are on the y-axis. Choose all that apply.

Starting at home, Luis traveled uphill to the gift store for 50 minutes at just 6 mph. He then traveled back home along the same

Georgia [21]

Answer:

Therefore his average speed for the entire trip is 8 mph.

Step-by-step explanation:

Given, In 50 minutes Luis traveled uphill to gift store at 6 mph.

6 mph means in 1 h= 60 minutes Luis can covered 6 mile.

In 1 minutes Luis can covered $\frac{6}{60}$ mile.

In 50 minutes Luis can covered $\frac{6\times 50}{60}$ mile

= 5 mile

Again when he come back at home, the speed was 12 mph.

12 mph means in 1 h= 60 minutes Luis can covered 12 mile.

Therefore he traveled 12 mile in 60 minutes

He traveled 1 mile in $\frac{60}{12}$ minutes.

He traveled 5 mile in $\frac{60\times 5}{12}$ minutes

= 25 minutes.

Total distance for the entire trip is =(5+5) mile=10 mile

Total time for the entire trip is = (50+25) minutes = 75 minutes

$\textrm{Average speed}=\frac{\textrm{Total distance}}{\Textrm {total time}}$

$=\frac{10}{75}$ m/min

$=\frac{10\times 60}{75}$ mph

= 8 mph

Therefore his average speed for the entire trip is 8 mph.

4 0

3 years ago

A person invested $8,400 in an account growing at a rate allowing the money to double every 11 years. How much money would be in

Salsk061 [2.6K]

Answer:

your on this app but yet a man has fallen into the in lego city

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A student has a class that is supposed to end at 9:00am and another that is supposed to begin at 9:10am. Suppose the actual endi

never [62]

Answer:

0.8340

Step-by-step explanation:

The aim of this question is to find the probability that the student makes it to the second class prior to when the lecture commences.

Provided that A, B, and C are independent of each other from the full complete question.

Then; we want P(A + C < B)

So A $\sim$ N (2, 1.5)

B $\sim$ N (10, 1)

C $\sim$ N (6, 1)

P(A + C - B < 0)

Since they are normally distributed( i.e. A + C - B)

Then;

E(A + C -B) = E(A) + E(C) - (B)

E(A + C -B) = 2 - 10 + 6

E(A + C -B) = -2

Var(A+C -B) = Var(A) + Var (B) + Var (C)

Var(A + C -B) = (1.5)² + (1)² + (1)²

Var(A + C -B) = 4.25

The standard deviation = $\sqrt{4.25}$

The standard deviation = 2.06155

$P(A+C-B) = p \Big (Z < \dfrac{0-(-2)}{2.06155} \Big )$

$P(A+C-B) = p \Big (Z \le 0.97014 \Big )$

$\mathbf{P(A+C-B) = 0.8340}$

7 0

3 years ago

Solving Quadratic Equations

NISA [10]

Answer:

a) x= 5 or x= -3

b) x = -4 + $\sqrt{26}$ or -4 - $\sqrt{26}$

Step-by-step explanation:

a) $x^{2}$ - 2x - 15 =0

x = $\frac{-(-2) +\sqrt{(-2)^{2} - 4(1)(-15)}}{2(1)}$ or $\frac{-(-2) -\sqrt{(-2)^{2} - 4(1)(-15)}}{2(1)}$

x= 5 or x=-3

b) $x^{2}$ + 8x - 10 =0

x = $\frac{-(8) +\sqrt{(8)^{2} - 4(1)(-10)}}{2(1)}$ or $\frac{-(8) -\sqrt{(8)^{2} - 4(1)(-10)}}{2(1)}$

x = -4 + $\sqrt{26}$ or -4 - $\sqrt{26}$

7 0

3 years ago

Find three consecutive intergers such that three times the third is 102 more than the sum of the first and second.

Pavlova-9 [17]

Answer:

A) x + x+1 + x+2

We don't need equation A)

B) 3*(x+2) = (x + X+1) +102

3x + 6 = 2x + 1 + 102

x = 97

Numbers are 97, 98 and 99

Step-by-step explanation:

6 0

3 years ago