Answer:
The answer is 7(x + 25)
Step-by-step explanation:
The question states 25 push-ups plus x amount of push-ups per day. To find how many push-ups per week, we multiply how many push-ups per day by 7 (a week).
Step-by-step explanation:
Answer:
2. Judy =$5
Ben= $4
Step-by-step explanation:
2. Let Judy = x and Ben = y
8x + 10y = 80
9x + 5y = 65
Solve these simultaneous equations.
8
x + 10
y = 80
18
x + 10
y = 130
Take the second equation away from the first equation
−
10
x = −
50
x = 5
This means that Judy gets paid $5 an hour.
Therefore, Ben gets paid $4 an hour.
Answer:
3. 24 quarters and 16 dimes
Step-by-step explanation:
3. Let the number of dimes = x and the number of quarters = y
Value Value
Type Number of of
of of EACH ALL
coin coins coin coins
-------------------------------------------
dimes x $0.10 $0.10x
quarters y $0.25 $0.25y
-------------------------------------------
TOTALS 40 $7.60
x + y = 40
0.10x + 0.25y = 7.60
Get rid of decimals by multiplying every term by 100:
10x + 25y = 760
So we have the system of equations:
x + y = 40
10x + 25y = 760
We solve by substitution. Solve the first equation for y:
x + y = 40
y = 40 - x
Substitute (40 - x) for y in 10x + 25y = 760
10x + 25(40 - x) = 760
10x + 1000 - 25x = 760
-15x + 1000 = 760
-15x = -240
x = 16 = the number of dimes.
Substitute in y = 40 - x
y = 40 - (16)
y = 24 quarters.
$1.60 + $6.00 = $7.60
I hope that helped, sorry for taking so long :-)
Its the same thing because your just switching the order which doesn't matter when your adding or multiplying
Answer:
Step-by-step explanation:
Researchers measured the data speeds for a particular smartphone carrier at 50 airports.
The highest speed measured was 76.6 Mbps.
n= 50
X[bar]= 17.95
S= 23.39
a. What is the difference between the carrier's highest data speed and the mean of all 50 data speeds?
If the highest speed is 76.6 and the sample mean is 17.95, the difference is 76.6-17.95= 58.65 Mbps
b. How many standard deviations is that [the difference found in part (a)]?
To know how many standard deviations is the max value apart from the sample mean, you have to divide the difference between those two values by the standard deviation
Dif/S= 58.65/23.39= 2.507 ≅ 2.51 Standard deviations
c. Convert the carrier's highest data speed to a z score.
The value is X= 76.6
Using the formula Z= (X - μ)/ δ= (76.6 - 17.95)/ 23.39= 2.51
d. If we consider data speeds that convert to z scores between minus−2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant?
The Z value corresponding to the highest data speed is 2.51, considerin that is greater than 2 you can assume that it is significant.
I hope it helps!
