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3 years ago

(5a +7b)b/(b+2b)=? 4a ,4ab , 2a+4b , 5a+7b/3 , 5a+5b

2 answers:

6 0

The correct answer is: [D]: " $\frac{5a + 7b}{3}$ " .
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Explanation:
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$\frac{b(5a + 7b)}{b +2b}$ ;

Simplify the "denominator" ; → (b + 2b = 1b + 2b = 3b) ;

and rewrite:

$\frac{b(5a + 7b)}{3b}$ ;

Cancel out the "b" in the "numerator" to a "1" ;
and cancel the "3b" in the "denominator to a "3" ;

{since: "b ÷ b = 1 "; and since: "3b ÷ b = 3" } ;

and rewrite as:
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$\frac{5a + 7b}{3}$ ;
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→ which is: Answer choice: [D]: " $\frac{5a + 7b}{3}$ " .
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avanturin [10]3 years ago

6 0

Answer:

Step-by-step explanation:

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insulin comes in 10 cubic centimeter(cc) vials labeled in the number of units of insulin per cubic centimeter of fluid. A vial m

Sergio [31]

Answer:

8.25 cc fluid should be drawn into the syringe from the U40 vial.

Step-by-step explanation:

The amount of a U40 vial insulin fluid is 10 cc.

A vial marked U40 has 40 units of insulin per cubic centimeter of fluid.

Suppose, if a patient needs 33 units of insulin, then $x$ cc fluid should be drawn.

So, according to the ratio of "cubic centimeter" to "the units of insulin", the equation will be.........

$\frac{x}{33}=\frac{10}{40}\\ \\40x=330\\ \\ x=\frac{330}{40}=8.25$

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4 years ago

30,840 in expanded form?

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3 years ago

Read 2 more answers

Heart failure is due to either natural occurrences (89%) or outside factors (11%). Outside factors are related to induced substa

Firdavs [7]

Answer:

a) 11% probability that the first patient with heart failure that enters the emergency room has the conditions due to outside factors.

b) 0.0871 = 8.71% probability that third patient with heart failure that enters the emergency room is the first one due to outside factors.

c) The mean number of heart failure patients with the condition due to natural causes that enter the emergency room before the first patient with heart failure from outside factors is 9.09.

Step-by-step explanation:

We have these following probabilities:

89% probability that a patient has heart failure due to natural occurrences.

11% probability that a patient has heart failure due to outside factors.

A) What is the probability that the first patient with heart failure that enters the emergency room has the conditions due to outside factors?

We assume that causes of heart failure between individuals are independent, which means that the probabilities are the same for each patient. So:

11% probability that the first patient with heart failure that enters the emergency room has the conditions due to outside factors.

B) What is the probability that third patient with heart failure that enters the emergency room is the first one due to outside factors?

First two due to natural occurrences, each with 89% = 0.89 probability.

Third due to outside factors, with 11% = 0.11 probability. So

0.89*0.89*0.11 = 0.0871

0.0871 = 8.71% probability that third patient with heart failure that enters the emergency room is the first one due to outside factors.

C) What is the mean number of heart failure patients with the condition due to natural causes that enter the emergency room before the first patient with heart failure from outside factors?

11% with outside factors, so the mean number of patients before the first with outside factors is given by

m = 1/0.11 = 9.09

The mean number of heart failure patients with the condition due to natural causes that enter the emergency room before the first patient with heart failure from outside factors is 9.09.

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3 years ago

Urn I contains three red chips and one white chip. Urn II contains two red chips and two white chips. One chip is drawn from eac

Lelechka [254]

Answer:

0.7

Step-by-step explanation:

The Probability of Chips in Urn 1

P(R) = 3/4

P(W) = 1/4

The Probability of Chips in Urn 2

P(R) = 1/2

P(W) = 1/2

The problem can be solved using probability tree method. The probability tree of the given question is attached with the answer.

The Probability tree shows the exchange of chips firstly and then the final probability of Red Chips and White Chips in Urn 1.

To find the final answer we can sum up the probabilities of the branches which are ticked in the diagram.

Probability of Branch 1 (ticked in diagram)

3/4 x 3/5 x 3/4 = 0.3375

Probability of Branch 2 (ticked in diagram)

3/4 x 2/5 x 2/4 = 0.15

Probability of Branch 3 (ticked in diagram)

1/4 x 2/5 x 1 = 0.1

Probability of Branch 4 (ticked in diagram)

1/4 x 3/5 x 3/4 = 0.1125

Final Probability = 0.3375 + 0.15 + 0.1 + 0.1125

= 0.7 Answer

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3 years ago

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qwelly [4]

45 is 36% of 125

Hope this helps!

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3 years ago