Okay so I have finals in a week on the picture above ^ How can I pass? (Asking for advice)

A taxi travels 25 km east of airport. Then it travels from that point to a point that is 40 km west of the airport. Finally the

Marizza181 [45]

The answer is: " 95 km " .
__________________________________________________
Explanation:
__________________________________________________

 ("airport")
• {Step 1): " 25 km east " •
│===============================> │

• {Step 2): " 40 km west ["of that point"] " ; • │<==========================================│

 • {Step 3: [back to airport] = " (40 km − 25 km) " ;

 • ("airport").
│=======>│
 _________________________________________________

 Note: To find "total distance traveled:

" 40 km + (40 km) + (40 km − 25 km) " ;

= " 40 km + (40 km) + (15 km) " ;

= " 40 km + 40 km + (15 km) " ;

= 80 km + 15 km ;

= 95 km .
____________________________________________________

The answer is: " 95 km " .
____________________________________________________

4 0

4 years ago

A cable company uses the formula S = 20,000 (1 + 0.08)y to estimate the number

vlada-n [284]

Answer:

C. Every year the number of subscribers is estimated to increase by 8% over the

number of subscribers the year before.

Step-by-step explanation:

Using the formula :

S = 20,000 (1 + 0.08)^y to estimate ;

S = number of subscribers ; y = years

The expression : (1 + 0.08)^y

0.08 = 0.08 * 100% = 8%

(1 + 0.08)^y represents a continous growth or increase in the number of subcsribers by 8% over the number of subscribers the preceeding year

4 0

3 years ago

Three cards are drawn from a standard deck of 52 cards without replacement. Find the probability that the first card is an ace,

MrRissso [65]

Answer:

$4.82\cdot 10^{-4}$

Step-by-step explanation:

In a deck of cart, we have:

a = 4 (aces)

t = 4 (three)

j = 4 (jacks)

And the total number of cards in the deck is

n = 52

So, the probability of drawing an ace as first cart is:

$p(a)=\frac{a}{n}=\frac{4}{52}=\frac{1}{13}=0.0769$

At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is

$n-1=51$

Therefore, the probability of drawing a three at the 2nd draw is

$p(t)=\frac{t}{n-1}=\frac{4}{51}=0.0784$

Then, at the third draw, the previous 2 cards are not replaced, so there are now

$n-2=50$

cards in the deck. So, the probability of drawing a jack is

$p(j)=\frac{j}{n-2}=\frac{4}{50}=0.08$

Therefore, the total probability of drawing an ace, a three and then a jack is:

$p(atj)=p(a)\cdot p(j) \cdot p(t)=0.0769\cdot 0.0784 \cdot 0.08 =4.82\cdot 10^{-4}$

4 0

3 years ago

I still don’t understand can someone give a helping hand?

Fofino [41]

Answer:

A

Step-by-step explanation:

A, B, C, and D, are all functions. the " { " is saying that those functions can are different at different points. For A the function is $A= -x-2$ when $x< 2$ and then $A=x+4$ when $x\geq -2$ . At the point $x=-2$ you can see that the function on the graph is going in two different directions but it stops at $x=-2$ .

Based on the graph you can see that the line on the left has a slope of -1 and if it kept going it would cross the y-axis at $y=-2$ , therefore the equation for the first line is $y =-x-2$ when $x$

The second line on the right side has a slope of 1 and crosses the y-axis at $y=4$ so the equation for that line is $y=x+4$ when $x\geq -2$ .

The only answer that matches these requirements is A

6 0

3 years ago

A company has budgeted 5 1/4 hours to complete a project, with 1/3 of the time spent on research. How much time does the company

elena-s [515]

Given that a company budgeted 5 1/4 hours to complete a project, determine how much time they spent on research if they spent 1/3 of the total budget.

First, convert the budget from hours into minutes.

5 1/4 hours = 315 minutes

1 hour = 60 min

1/4 hour = 15 min

60 x 5 = 300

300 + 15 = 315

Then, divide the minutes by 3 or multiply it by 1/3.

315 / 3 = 105

315 x 1/3 = 105

Lastly convert to a mixed number.

1 3/4 hour

Thus, the company plans to spend 1 3/4 hours or 1 hour and 45 minutes on research.

7 0

3 years ago