Hi Lala4eva54,
Your Question:
A total of 123 fifth-grade students are going to ford Verde state historic park each bus holds 38 students all of the buses are full Except one how many students will be in the bus that is not feel.
Solution:
123 / 38 = 3.236.....
38 x 3 = 114
123 - 114 = 9
Answer:
There will be 9 students in the bus thats not full.
Answer:
30 Hours
Step-by-step explanation:
8h=240
240÷8=30
Let's check. 6 one-dollars = $6 15 five-dollars = $75 9 ten-dollars = $90 Add them all up to get $171, so that is correct. Add the number of one-dollar bills and the number of ten-dollar bills together. 6 + 9 = 15, which is the number of five-dollar bills, so that is correct as well. Add all the numbers of bills together, 6 + 9 + 15 = 15 + 15 = 30, so that is correct
The equation most closely models the line of best fit for the scatter plot is y = 2.5x
<h3>Equation of a line</h3>
A line is the distance between two points. The equation of a line in point-slope form is given as y =mx + b
- m is the slope
- b is the intercept
Using the coordinate points on the line (3, 5000) and (9, 20000)
Get the slope
Slope = 20000-5000/9-3
Slope = 15000/6 = 2500
Determine the y-intercept
5000 = 2500(3) + b
b = 5000 - 7500
b = -2500
Determine the equation
y = 2.5x - 2.5
Hence the equation most closely models the line of best fit for the scatter plot is y = 2.5x
Learn more on line plot here: brainly.com/question/8989301
#SPJ1
Answer:
(a)3
(b)4
(c)6
(d)5
Step-by-step explanation:
(a)Josie rolls a six-sided die 18 times.
P(she rolls a two)=1/6
Therefore, the estimated number of times she rolls a two in 18 trials
=1/6 X 18
=3
(b)Slips of paper are numbered 1 through 10.
P(the number 10 appear)=1/10
If one slip is drawn and replaced 40 times, expected number of 10
=1/10 X 40
=4
(c)A spinner consists of 10 equal- sized spaces: 2 red, 3 black, and 5 white.
P(red)=2/10
If the spinner is spun 30 times, expected number of red space
=2/10 X 30
=6
(d)A card is picked from a standard deck
P(drawing an ace)=4/52
If the card is picked 65 times and replaced each time.
Expected Number of Aces =4/52 X 65 =5
