Since c=15 and d=3, we replace the c and d variables with 15 and 3.
0.5x15-1.7x3
0.5x15=7.5
1.7x3=5.1
Now subtract.
7.5-5.1=2.4
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hope it helps
Answer:
if you are lookng for C it is 17
Step-by-step explanation:
Answer:
A. No solution
Step-by-step explanation:
Choose 1 answer:
A. No solutions
B. Exactly one solution
C. Infinitely many solutions
Solution
Given:
4(y-30)=4y+12
Open parenthesis
4y-120=4y+12
Collect like terms
4y-4y=12+120
0=142
There is no solution to the equation, therefore, the answer is A
Answer:
the answer is 160 that the answer
Answer:
68% of pregnancies last between 250 and 282 days
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 266
Standard deviation = 16
What percentage of pregnancies last between 250 and 282 days?
250 = 266 - 16
250 is one standard deviation below the mean
282 = 266 + 16
282 is one standard deviation above the mean
By the Empirical Rule, 68% of pregnancies last between 250 and 282 days
