1)
Let the number of fish sandwiches sold br represented by x
Then the number of grilled cheese sold is 2X
and the number of cheeseburgers 3(2x)
x + 2x + 3(2x) = 225
and solve
2)
Let x represent the no. of pounds of peanuts
Then x + 14 represents the no of pounds of the mixture
And
(2.25)x + (3.25)(14) = (2.65)(x + 14
and solve
3) Try on your own
plz mark me as brainliest :)
Given the inequality expression:
w−0.8>−2.4
Inequalities are expressions separated by an unequal sign.
According to the given question, we need to find the solution for "w"
w−0.8>−2.4
Add 0.8 to both sides
w−0.8+0.8>−2.4 + 0.8
w > -2.4 + 0.8
Simplify the result by adding to have:
w > -1.6
Find the number line representing the solution attached;
Learn more on number line here: brainly.com/question/24644930
Answer: The sum of the measures of the angle inside of a triangle add up to 180 degrees. That's all we're doing over here. And so let's see if we can simplify this a little bit. So these two guys-- 90 plus 90's going to be 180, so you get 180 minus theta …
Author: Sal Khan
Step-by-step explanation:
Answer:
Five million four thousand three hundred
Step-by-step explanation:
Answer:
a. 2.28%
b. 30.85%
c. 628.16
d. 474.67
Step-by-step explanation:
For a given value x, the related z-score is computed as z = (x-500)/100.
a. The z-score related to 700 is (700-500)/100 = 2, and P(Z > 2) = 0.0228 (2.28%)
b. The z-score related to 550 is (550-500)/100 = 0.5, and P(Z > 0.5) = 0.3085 (30.85%)
c. We are looking for a value b such that P(Z > b) = 0.1, i.e., b is the 90th quantile of the standard normal distribution, so, b = 1.281552. Therefore, P((X-500)/100 > 1.281552) = 0.1, equivalently P(X > 500 + 100(1.281552)) = 0.1 and the minimun SAT score needed to be in the highest 10% of the population is 628.1552
d. We are looking for a value c such that P(Z > c) = 0.6, i.e., c is the 40th quantile of the standard normal distribution, so, c = -0.2533471. Therefore, P((X-500)/100 > -0.2533471) = 0.6, equivalently P(X > 500 + 100(-0.2533471)), and the minimun SAT score needed to be accepted is 474.6653
