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3 years ago

The price of parking for 2 hours or less at the airport is increasing

Mathematics

2 answers:

maria [59]3 years ago

8 0

Answer:

50% increase

Step-by-step explanation:

half of 5.00 is 2.50

fredd [130]3 years ago

5 0

Answer:

it increased 50%

Step-by-step explanation:

5 / 2.5 = 2

100 / 2 = 50%

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HELP ME ASAP!!!!!!!!

Marina CMI [18]

Answer:

Step-by-step explanation:

1) 30 : 100 = x : 299

x = (30 x 299)/100

x = 89,70 $

2) 299 - 89,70 = 209,30 $

3) 5 : 100 = x : 209,30

x = 10,465 $

209,30 + 10,465 = 219,77 $

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3 years ago

What is X>195 on a number line

zlopas [31]

Step-by-step explanation:

anything greater than 195.

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3 years ago

Read 2 more answers

Write the next three terms of the arithmetic sequence.<br> First term: 2<br> Common difference: 13

IRINA_888 [86]

15, 28, 41.

add the common difference to each consecutive term to find the next term in the sequence.

2+ 13= 15
15+ 13= 28
28+ 13= 41

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3 years ago

Please please please help me with this...

dlinn [17]

Answer:

It's option d.

Step-by-step explanation:

The line y = x + 2 has domain x < 2 (because of the clear circle.)

The lines y = x + 1 has domain x ≥ 2. ( because of the filled circle).

7 0

3 years ago

A raffle offers one $8000.00 prize, one $4000.00 prize, and five $1600.00 prizes. There are 5000 tickets sold at $5 each. Find t

Harman [31]

Answer:

The expectation is $E(1 )= -\$ 1$

Step-by-step explanation:

From the question we are told that

The first offer is $x_1 = \$ 8000$

The second offer is $x_2 = \$ 4000$

The third offer is $\$ 1600$

The number of tickets is $n = 5000$

The price of each ticket is $p= \$ 5$

Generally expectation is mathematically represented as

$E(x)=\sum x * P(X = x )$

$P(X = x_1 ) = \frac{1}{5000}$ given that they just offer one

$P(X = x_1 ) = 0.0002$

Now

$P(X = x_2 ) = \frac{1}{5000}$ given that they just offer one

$P(X = x_2 ) = 0.0002$

Now

$P(X = x_3 ) = \frac{5}{5000}$ given that they offer five

$P(X = x_3 ) = 0.001$

Hence the expectation is evaluated as

$E(x)=8000 * 0.0002 + 4000 * 0.0002 + 1600 * 0.001$

$E(x)=\$ 4$

Now given that the price for a ticket is $\$ 5$

The actual expectation when price of ticket has been removed is

$E(1 )= 4- 5$

$E(1 )= -\$ 1$

4 0

2 years ago