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3 years ago

Ticket costs £14.50 add 7% increase in price

2 answers:

8 0

Solutions

We know that a ticket costs £14.50 and it is increased by 7% in price. To solve the problem lets use variables.

Let the Variable Y = 7% of 14.5

Equation: Y = P% × (X)

Solve for our equation Y

Y = P% × (X)

Y = 7% × 14.5

Now lets convert the percent to a decimal:

Percent = Out of 100

Percent = 7% / 100 = 0.07

Y = 0.07 × 14.5

Y = 1.015

SCORPION-xisa [38]3 years ago

3 0

If you would like to know the cost of the ticket after the 7% increase, you can calculate this using the following steps:

7% of £14.50 = 7% * 14.50 = 7/100 * 14.50 = £1.015

£14.50 + £1.015 = £15.515

The final cost of the ticket would be £15.515.

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Who runs the fastest ? (ONLY ANSWER IF YOU KNOW IT BECAUSE CORRECT ONE GETS BRAINLIEST)

Genrish500 [490]

Answer:

Stephanie runs the fastest per second: 14 feet per second

Step-by-step explanation:

To find the fastest speed, you need to see which person runs the most per second:

Stephanie runs 14 feet per second

Brooke runs 589 feet in 44 seconds.

Find the amount of feet per second by dividing:

$\frac{589}{44} =13.4$

Brooke runs about 13.4 feet per second

Will runs 1 mile in 454 seconds.

Convert the miles to feet:

There are 5,280 feet in a mile. If Will runs 5,280 feet in 454 seconds, divide to find how much he runs per second:

$\frac{5280}{454}= 11.6$

Will runs about 11.6 feet per second.

Rob runs 548 feet in a minute.

Convert the minute into seconds. There are 60 seconds in a minute. Divide the feet by the seconds to find how fast he runs per second:

$\frac{548}{60}= 9.1$

Rob runs about 9.1 feet per second.

Stephanie runs 14 feet per second, the fastest time:

| 14 |, 13.4, 11.6, 9.1

Finito.

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3 years ago

In a lab, a colony of 100 bacteria is placed on a petri dish. The population triples every hour.

Evgesh-ka [11]

Answer:

a). 8100 bacteria

b). 519 bacteria

c). Infinite

Step-by-step explanation:

Population growth of a bacteria is given by the exponential function,

f(x) = $P_{0}(1+\frac{r}{100})^{t}$

Where f(x) = Population after time 't'

$P_{0}$ = Initial population

r = Growth rate

t = Duration after t hours

If "100 bacteria gets tripled every hour"

300 = $100(1+\frac{r}{100})^{1}$

3 = 1 + $\frac{r}{100}$

r = 100×(3 - 1)

r = 200

So the function is, f(x) = $100(3)^{t}$

a). Population after 4 hours,

f(4) = $100(3)^{4}$

= 8100 bacteria

b). Population after 90 minutes Or 1.5 hours

f(1.5) = $100(3)^{1.5}$

= 519.61

≈ 519 bacteria

c). Population after 72 hours,

f(72) = $100(3)^{72}$

= Infinite

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3 years ago

Point J is the midpoint of segment FG with endpoints F(1,4) and G(5,12). Point K is the midpoint of segment GH with endpoints G(

faltersainse [42]

Answer:

1 unit

Step-by-step explanation:

If point J is the midpoint of segment FG with endpoints F(1,4) and G(5,12), then its coordinates are

$x_J=\dfrac{x_F+x_G}{2}=\dfrac{1+5}{2}=3\\ \\y_J=\dfrac{y_F+y_G}{2}=\dfrac{4+12}{2}=8$

If point K is the midpoint of segment GH with endpoints G(5,12) and H(-1,4), then its coordinates are

$x_K=\dfrac{x_G+x_H}{2}=\dfrac{5+(-1)}{2}=2\\ \\y_K=\dfrac{y_G+y_H}{2}=\dfrac{12+4}{2}=8$

The measure of JK is

$JK=\sqrt{(3-2)^2+(8-8)^2}=\sqrt{1^2+0^2}=1\ unit$

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3 years ago

What is the total value of the digit 5 in the number 57,634

melomori [17]

Hey there!

$\bold{57,634}$

$\bold{5:}$ $\bold{ten}$ $\bold{thousands}$
$\bold{7:}$ $\bold{thousands}$
$\bold{6:}$ $\bold{hundreds}$
$\bold{3:}$ $\bold{tens}$
$\bold{4:}$ $\bold{ones}$
$\boxed{\boxed{\bold{Answer:ten-thousand(s)}}}$

Good luck on your assignment and enjoy your day!

~ $\bold{LoveYourselfFirst:)}$

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2 years ago

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