Answer:
0.9586
Step-by-step explanation:
From the information given:
7 children out of every 1000 children suffer from DIPG
A screening test designed contains 98% sensitivity & 84% specificity.
Now, from above:
The probability that the children have DIPG is:


= (0.98 × 0.007) + 0.16( 1 - 0.007)
= 0.16574
So, the probability of not having DIPG now is:



= 0.9586
Answer:
Step-by-step explanation:
Given that 33.4% of people have sleepwalked.
Sample size n =1459
Sample favourable persons = 526
Sample proportion p = 
Sample proportion p is normal for large samples with mean = 0.334 and
std error = 
a) P(526 or more of the 1459 adults have sleepwalked.)

b) Yes, because hardly 1.4% is the probability
c) 33.4 is very less compared to the average. Either sample should be improved representing the population or population mean should be increased accordingly.
Answer: 8
Explanation: Without the discount, 8 frozen yogurts for $4 will come to $32. Deduct $2 from the total cost (32-2), and now they're paying $30.
<span>1.5
In C or C++, a floating point literal is nothing more than a sequence of decimal digits with either or both an decimal point, or an exponent with optionally a type specifier. The value could be any of the following: 1.5, 1.5e0, 0.15e1, 15e-1, etc.</span>