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Mnenie [13.5K]
3 years ago
8

Find the Volume (nearest integer):*

Mathematics
1 answer:
Illusion [34]3 years ago
3 0

Answer:

the correct answer is 4) 1692

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Antonio has wood that is 4 1/2 feet. And he wants to cut it into 0.75 foot sections. How many equal sections of wood he will he
g100num [7]

Answer:

6

Step-by-step explanation:

4 1/2 / 0.75 = 4.5 / 0.75 = 6

8 0
3 years ago
How many grams of active ingredient is contained in 500ml of a 30% solution
netineya [11]

let's reword it, since the wording is crummy

"How many grams of active ingredient is contained in 500mL of a solution which is 30% of the active ingredient?"

namely, what is 30% of 500?

\begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{30\% of 500}}{\left( \cfrac{30}{100} \right)500}\implies 150~mL

4 0
2 years ago
PLZZZ HELP NOW!!!<br> BRAINLIESTTT<br> Images below
Elden [556K]

Answer:

B

Step-by-step explanation:

pretty sure it's right

8 0
3 years ago
Read 2 more answers
In a start-up company which has 20 computers, some of the computers are infected with virus. The probability that a computer is
Alex777 [14]

Answer:

(1) The probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.

(2) The probability at least 5 computers are infected is 0.949.

Step-by-step explanation:

The probability that a computer is defective is, <em>p</em> = 0.40.

(1)

Let <em>X</em> = number of computers to be tested before the 1st defect is found.

Then the random variable X\sim Geo(p).

The probability function of a Geometric distribution for <em>k</em> failures before the 1st success is:

P (X = k)=(1-p)^{k}p;\ k=0, 1, 2, 3,...

Compute the probability that the technician tests at least 5 computers before the 1st defective computer is found as follows:

P (X ≥ 5) = 1 - P (X < 5)

              = 1 - [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)]

              =1 -[(1-0.40)^{0}0.40+(1-0.40)^{1}0.40+(1-0.40)^{2}0.40\\+(1-0.40)^{3}0.40+(1-0.40)^{4}0.40]\\=1-[0.40+0.24+0.144+0.0864+0.05184]\\=0.07776\\\approx0.078

Thus, the probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.

(2)

Let <em>Y</em> = number of computers infected.

The number of computers in the company is, <em>n</em> = 20.

Then the random variable Y\sim Bin(20,0.40).

The probability function of a binomial distribution is:

P(Y=y)={n\choose y}p^{y}(1-p)^{n-y};\ y=0,1,2,...

Compute the probability at least 5 computers are infected as follows:

P (Y ≥ 5) = 1 - P (Y < 5)

             = 1 - [P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3) + P (Y = 4)]               =1-[{20\choose 0}(0.40)^{0}(1-0.40)^{20-0}+{20\choose 1}(0.40)^{1}(1-0.40)^{20-1}\\+{20\choose 2}(0.40)^{2}(1-0.40)^{20-2}+{20\choose 3}(0.40)^{3}(1-0.40)^{20-3}\\+{20\choose 4}(0.40)^{4}(1-0.40)^{20-4}]\\=1-[0.00004+0.00049+0.00309+0.01235+0.03499]\\=1-0.05096\\=0.94904

Thus, the probability at least 5 computers are infected is 0.949.

7 0
3 years ago
Mrs. Arthur's fourth grade class has Art on Monday. They go to computer class every Tuesday and Thursday. Mrs. Arthur's class ha
Nutka1998 [239]
                  Monday  Tuesday    Wednesday    Thursday       Friday
 week1:       Art           computer  music               computer       gym
  week2:      Art           computer   music               Computer      gym




4 0
3 years ago
Read 2 more answers
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