Answer:
$600
Step-by-step explanation:
Let "x" represent their total budget for the film.
Amount spent on costumes = $330
Percent spent on costumes = 55%
Therefore:
55% of x = $330
55/100 × x = 330
55x/100 = 330
Cross multiply
55x = 100 × 330
55x = 33,000
Divide both sides by 55
x = 33,000/55
x = 600
Total budget = x = $600
Answer:
13/15
Step-by-step explanation:
2/3 + 1/5
You first,
- Find the L. C. M, i hope u know the full name, if u don't, let me know in the comment section.
- the L. C. M is 15
- therefore it will be 10+3/15
- 13/15
Complete question:
He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is
a) less than 8 minutes
b) between 8 and 9 minutes
c) less than 7.5 minutes
Answer:
a) 0.0708
b) 0.9291
c) 0.0000
Step-by-step explanation:
Given:
n = 47
u = 8.3 mins
s.d = 1.4 mins
a) Less than 8 minutes:
P(X' < 8) = P(Z< - 1.47)
Using the normal distribution table:
NORMSDIST(-1.47)
= 0.0708
b) between 8 and 9 minutes:
P(8< X' <9) =![[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]](https://tex.z-dn.net/?f=%20%5B%5Cfrac%7B8-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%3C%20%5Cfrac%7BX%27-u%7D%7Bs.d%2F%20%5Csqrt%7Bn%7D%7D%20%3C%20%5Cfrac%7B9-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%5D)
= P(-1.47 <Z< 6.366)
= P( Z< 6.366) - P(Z< -1.47)
Using normal distribution table,
0.9999 - 0.0708
= 0.9291
c) Less than 7.5 minutes:
P(X'<7.5) = ![P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]](https://tex.z-dn.net/?f=%20P%20%5BZ%3C%20%5Cfrac%7B7.5-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%5D%20)
P(X' < 7.5) = P(Z< -3.92)
NORMSDIST (-3.92)
= 0.0000
Answer:
(3x+4)(5x+7)
Step-by-step explanation:
15x^2
+41x+28
Factor the expression by grouping. First, the expression needs to be rewritten as 15x^2
+ax+bx+28. To find a and b, set up a system to be solved.
a+b=41
ab=15×28=420
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 420.
1,420
2,210
3,140
4,105
5,84
6,70
7,60
10,42
12,35
14,30
15,28
20,21
Calculate the sum for each pair.
1+420=421
2+210=212
3+140=143
4+105=109
5+84=89
6+70=76
7+60=67
10+42=52
12+35=47
14+30=44
15+28=43
20+21=41
The solution is the pair that gives sum 41.
a=20
b=21
Rewrite 15x^2
+41x+28 as (15x^2
+20x)+(21x+28).
(15x^2
+20x)+(21x+28)
Factor out 5x in the first and 7 in the second group.
5x(3x+4)+7(3x+4)
Factor out common term 3x+4 by using distributive property.
(3x+4)(5x+7)
Answer:
(-8/7 ; 5/7)
Step-by-step explanation:
5t + 1/2s = 3 - - - (1)
3t - 6s = 9 - - - - - (2)
Multiply (1) by 12 and (2) by 1
Add the result to eliminate s
60t + 6s = 36
3t - 6s = 9
____________
63t = 45
t = 45 / 63
t = 5/7
Put t = 5/7 in either (1) or (2) to obtain the value of s
3(5/7) - 6s = 9
15/7 - 6s = 9
-6s = 9 - 15/7
-6s = (63 - 15)/7
-6s = 48/7
s = 48/7 * - 1/6
s = - 8/7
