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2 years ago

a culture started with 5000 bacteria after 2 hours it grew to 6500 predict how mabacteria will be present after 18 hours

2 answers:

8 0

In 18 hour it grew 13,500, because it grew 1,500 in 2 hour and if you divide 18 by 2 it is 9 so 1,500 time 2 is 13,500.
hopefully I am correct and this helps you.

Bess [88]2 years ago

5 0

In 18 hours there are a total of 13,500 bacteria.

First, you'd find the difference between 6,500 and 5,000 (6,500 - 5,000 = 1,500). Next you'd divide 1,500 by 2 to find how many grow per hour (1,500/2 = 750). Finally, you multiply 750 by 18, to show how many will be present in 18 hours.

I hope this helps!

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2 years ago

What is the tenth digit in the quotient 120 ÷ 11

galina1969 [7]

20 is the tenth digit in the quotient

4 0

3 years ago

Read 2 more answers

Question 1 (1 point)

sergey [27]

Answer:

Step-by-step explanation:

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4 0

3 years ago

Consider the functions. F(x)=(x+1)2-4 and g(x)=-4|x+1| which statement compares the range of the functions?

Andreyy89

Answer:

<h2> The fourth</h2>

Step-by-step explanation:

Vertex of f is (-1, -4) so its range is limited to y≥-4

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3 0

3 years ago

For a binomial distribution with p = 0.20 and n = 100, what is the probability of obtaining a score less than or equal to x = 12

notsponge [240]

The binomial distribution is given by,
P(X=x) = $(^{n}C_{x})p^{x} q^{n-x}$
q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability <span>of obtaining a score less than or equal to 12.
</span>∴ P(X≤12) = $(^{100}C_{x})(0.2)^{x} (0.8)^{100-x}$
where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12
∴ P(X≤12) = $(^{100}C_{0})(0.2)^{0} (0.8)^{100-0} + (^{100}C_{1})(0.2)^{1} (0.8)^{100-1}$ + $(^{100}C_{2})(0.2)^{2} (0.8)^{100-2} + (^{100}C_{3})(0.2)^{3} (0.8)^{100-3}$ + $(^{100}C_{4})(0.2)^{4} (0.8)^{100-4} + (^{100}C_{5})(0.2)^{5} (0.8)^{100-5}$ + $(^{100}C_{6})(0.2)^{6} (0.8)^{100-6} + (^{100}C_{7})(0.2)^{7} (0.8)^{100-7}$ + $(^{100}C_{8})(0.2)^{8} (0.8)^{100-8} + (^{100}C_{9})(0.2)^{9} (0.8)^{100-9}$ + $(^{100}C_{10})(0.2)^{10} (0.8)^{100-10} + (^{100}C_{11})(0.2)^{11} (0.8)^{100-11}$ + $(^{100}C_{12})(0.2)^{12} (0.8)^{100-12}$

Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability.

7 0

3 years ago