C = children
A = adults
293 = c + a
1.50c + 2.50a = 676.50
We can use substitution to solve:
c + a = 293 subtract a to get c = 293 - a
Plug this into the second equation:
1.50(293 - a) + 2.50a = 676.50
439.5 - 1.50a + 2.50a = 676.50
439.5 + 1a = 676.50
a = 237
Substitute this into the first equation:
293 = c + 237
56 = c
Answer:
a: 0.9544 9 within 8 units)
b: 0.9940
Step-by-step explanation:
We have µ = 300 and σ = 40. The sample size, n = 100.
For the sample to be within 8 units of the population mean, we would have sample values of 292 and 308, so we want to find:
P(292 < x < 308).
We need to find the z-scores that correspond to these values using the given data. See attached photo 1 for the calculation of these scores.
We have P(292 < x < 308) = 0.9544
Next we want the probability of the sample mean to be within 11 units of the population mean, so we want the values from 289 to 311. We want to find
P(289 < x < 311)
We need to find the z-scores that correspond to these values. See photo 2 for the calculation of these scores.
We have P(289 < x < 311) = 0.9940
If you are talking about the area and perimeter of one of the triangular faces...
8 x 1.2 = 9.6
6 x 1.2 = 7.2
Perimeter = 9.6 + 7.2 + 7.2 = 24
Area = Linear Scale Factor²
1.2² = 1.44
7 x 1.44 = 10.08 (units)²
Answer=7/20
If there’s 20 total and there’s 7 blue it should be 7/20
Answer:
<em>Student 3 had the greatest percentage error.</em>
Step-by-step explanation:
<u>Student 1</u>
Actual length = 50in
Measured length = 49in
Error = 50 - 49 = 1
Error% = 1/50 × 100
= 1 × 2
= 2%
<u>Student</u><u> </u><u>2</u>
Actual length = 100cm
Measured length = 110cm
Error = 100 - 110 = 10
Error% = 10/100 × 100
= 10 × 1
= 10%
<u>Student</u><u> </u><u>3</u>
Actual length = 2cm
Measured length = 2.5cm
Error = 2 - 2.5 = 0.5
Error% = 0.5/2 × 100
= 0.5 × 50
= 25%
<u>Student</u><u> </u><u>4</u>
Actual length = 10ft
Measured length = 9½ft
Error = 10 - 9½ = ½
Error% = ½/10 × 100
= ½ × 10
= 1 × 5
= 5%
#teamtrees#WAP
