Jen has 54 pages left to read to meet her assignment requirement.
The inequality that represents this situation is p ≥ 85 - 31
To find how many pages Jen has left to read, we can subtract the number of pages she has already read from the minimum number of pages she needs to read.
The minimum number of pages Jen needs to read is 85, and she has already read 31 pages. So, the number of pages she has left to read, p, can be found by:
p = 85 - 31
p = 54
Therefore, Jen has 54 pages left to read.
To represent this situation with an inequality, we can use:
p ≥ 85 - 31
This inequality states that the number of pages Jen still needs to read, p, must be greater than or equal to the difference between the minimum number of pages she needs to read (85) and the number of pages she has already read (31).
Solving for p:
p ≥ 85 - 31
p ≥ 54
This means that Jen must read at least 54 more pages to meet her assignment requirement.
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we assume there is sometimes sunny days and sometimes rainy days, and on day 1, which we're going to call d1, the probability of sunny is 0.9. and then let's assume that a sunny day follows a sunny day with 0.8 chance, and a sunny day follows a rainy day with 0.6 chance. so, what are the chances that d2 is sunny?
Probability of D2 being sunny = 0.78.
On day 1, which is called D1, the probability of sunny is 0.9. It is also given that a sunny day follows a sunny day with 0.8 chance, and a sunny day follows a rainy day with 0.6 chance.
Therefore, we need to find the chances that D2 is sunny.
There are two possibilities for D2: either it can be a sunny day, or it can be a rainy day.
Now, Let us find the probability of D2 being sunny.
We have the following possible cases for D2.
D1 = Sunny; D2 = Sunny
D1 = Sunny; D2 = Rainy
D1 = Rainy; D2 = Sunny
D1 = Rainy; D2 = Rainy
The probability of D1 being sunny is 0.9.
When a sunny day follows a sunny day, the probability is 0.8.
When a sunny day follows a rainy day, the probability is 0.6.
Therefore, the probability of D2 being sunny is given by the formula:
Probability of D2 being sunny = (0.9 × 0.8) + (0.1 × 0.6) = 0.72 + 0.06 = 0.78.
Therefore, the probability that D2 is sunny are 0.78 or 78%.
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What is the Smallest Positive Integer with at least 8 odd Factors and at least 16 even Factors?
Therefore, the smallest positive integer with at least 8 odd factors and at least 16 even factors is N = 1800.
what is Combination?In mathematics, combination is a way to count the number of possible selections of k objects from a set of n distinct objects, without regard to the order in which they are selected.
The number of combinations of k objects from a set of n objects is denoted by [tex]nCk[/tex] or [tex]C(n,k),[/tex] and is given by the formula:
[tex]nCk = n! / (k! *(n-k)!)[/tex]
where n! denotes the factorial of n, i.e., the product of all positive integers up to n.
by the question.
Now, let's consider the parity (evenness or oddness) of the factors of N. A factor of N is odd if and only if it has an odd number of factors of each odd prime factor of N. Similarly, a factor of N is even if and only if it has an even number of factors of each odd prime factor of N. Therefore, the condition that N has at least 8 odd factors and at least 16 even factors can be expressed as:
[tex](a_{1} +1) * (a_{2} +1) * ... * (an+1) = 8 * 2^{16}[/tex]
Let's consider the factor 2 separately. Since N has at least 16 even factors, it must have at least 16 factors of 2. Therefore, we have a_i >= 4 for at least one prime factor p_i=2. Let's assume without loss of generality that p[tex]1=2[/tex] and [tex]a1 > =4.[/tex]
Now, let's consider the remaining prime factors of N. Since N has at least 8 odd factors, it must have at least 8 factors that are not divisible by 2. Therefore, the product (a2+1) * ... * (an+1) must be at least 8. Let's assume without loss of generality that n>=2 (i.e., N has at least three distinct prime factors).
Since a_i >= 4 for i=1, we have:
[tex]N > = 2^4 * p2 * p3 > = 2^4 * 3 * 5 = 240[/tex]
Let's now try to find the smallest such N. To minimize N, we want to make the product (a2+1) * ... * (an+1) as small as possible. Since 8 = 2 * 2 * 2, we can try to distribute the factors 2, 2, 2 among the factors (a2+1), (a3+1), (a4+1) in such a way that their product is minimized. The only possibility is:
[tex](a2+1) = 2^2, (a3+1) = 2^1, (a4+1) = 2^1[/tex]
This gives us:
[tex]N = 2^4 * 3^2 * 5^2 = 1800[/tex]
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