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Sphinxa [80]
2 years ago
5

Regular admission price for a family to enter an amusement park is $345.80. The Patel family used a discount coupon that gave th

em 35% off the normal price. How much did the family save?
I need help, I really would appreciate it.
Mathematics
1 answer:
mylen [45]2 years ago
3 0
100-35=65
65 over 100 x 345.80
=224.77
345.80-224.77=121.03
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Pls help <br> -21 ÷ b = 3, b = ? and uhm this one too<br> 3c + 1 = 10, c = ?
AfilCa [17]

Answer:

-21 ÷ b = 3, b =        b=-7

3c + 1 = 10, c =        c=1/7

Hope this helps!

5 0
3 years ago
As cars drive up a ramp at a multi-storey car park they go up 2 meters. the length of the ramp is 10 meters. find the angle betw
daser333 [38]
This is a right angle triangle problem
drawing a vertical line at from the point where the ramp touches the car park leaves a right angle triangle with the
opposite being 2m
hypothenus being 10m
adjacent unknown
we could use sine
SineO equal to opposite over hypothenus
SineO equal to 2/10
SineO equal to 0.2
O equal to Sine^1(0.2)
O equal to 11 .5
The angle between the ramp and the horizontal is 11.5 degrees
7 0
3 years ago
Evaluate the expression below when<br> X = 4 and y = (-3)<br> 4x + 2xy
zepelin [54]

Step-by-step explanation:

4x + 2xy=

4×4 + 2×4×-3

= 16 + (-24)

= - 8

7 0
3 years ago
What is the slope of the line graphed below
Solnce55 [7]
Slope is 1/3

Equation of the line is:

y = 1/3x - 8
6 0
3 years ago
For a certain​ candy, 15​% of the pieces are​ yellow, 1010​% are​ red, 2020​% are​ blue, 55​% are​ green, and the rest are brown
MatroZZZ [7]
A) the probability it is brown would be 50%; the probability it is yellow or blue would be 35%; the probability it is not green is 95%; the probability it is striped is 0%.
B) the probability of all brown would be 12.5%; the probability that the third one is the first red one drawn is 8.1%; the probability that none are yellow is 61.4%; the probability that at least one is green is 14.3%.

Explanation:
A) The probability that it is brown is the percentage of brown we have.  Brown is not listed, so we subtract what we are given from 100%:
100-(15+10+20+5) = 100-(50) = 50%.  The probability that one drawn is yellow or blue would be the two percentages added together:  15+20 = 35%.  The probability that it is not green would be the percentage of green subtracted from 100:  100-5=95%.  Since there are no striped candies listed, the probability is 0%.
B) Since we have an infinite supply of candy, we will treat these as independent events.  All 3 being brown is found by taking the probability that one is brown and multiplying it 3 times:
0.5*0.5*0.5 = 0.125 = 12.5%.  
To find the probability that the first one that is red is the third one drawn, we take the probability that it is NOT red, 100-10 = 90% = 0.9, for the first two, and the probability that it IS red, 10% = 0.1, for the last:
0.9*0.9*0.1 = 0.081 = 8.1%.

The probability that none are yellow is found by raising the probability that the first one is not yellow, 100-15=85%=0.85, to the third power:

0.85^3 = 0.614 = 61.4%.

The probability that at least one is green is computed by subtracting 1-(probability of no green).  We first find the probability that all three are NOT green:
0.95^3 = 0.857375
1-0.857375 = 0.143 = 14.3%.
3 0
3 years ago
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