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Lilit [14]
2 years ago
5

A student repeats the process of taking blocks out of a bag and replacing them 100 times. A green block is drawn 67 times. What

is a good
estimate for the probability of drawing out a green block from the bag?
P(green)
Mathematics
1 answer:
Alexeev081 [22]2 years ago
5 0

Using it's concept, it is found that a good estimate for the probability of drawing out a green block from the bag is of 0.67 = 67%.

<h3>What is a probability?</h3>

A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

In this problem, to get a good estimate, we get the probability taking the outcomes from the sample, that is, 67 green blocks out of 100 blocks, hence:

p = 67/100 = 0.67.

A good estimate for the probability of drawing out a green block from the bag is of 0.67 = 67%.

More can be learned about probabilities at brainly.com/question/14398287

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3 years ago
The pyramid shown has a square base that is 24 inches on each side. The slant height is 20 inches. What is the surface area of t
pashok25 [27]
The surface area for this is 1536 units.
4 0
3 years ago
15=x/4 - 3 how do i solve
zhuklara [117]

Answer: the answer is x=72

Step-by-step explanation:

Step 1: simplify both sides we know 15=x/4 -3 (the x holds a invisible 1)

the equation should look something Like: 15=1/4x-3 then FLIP the equation such as:

1/4x-3=15 then add 3 on both sides so 15+3=18

step 2: our equation is now 1/4x=18  (we see 4 is being divided the opposite of division is multiply) therefor multiply the 18 with 4

18*4=72 which means x=72

6 0
2 years ago
A multiple-choice standard test contains total of 25 questions, each with four answers. Assume that a student just guesses on ea
weqwewe [10]

Answer:

a)  8*88*10⁻⁶ ( 0.00088 %)

b) 0.2137 (21.37%)

Step-by-step explanation:

if the test contains 25 questions and each questions is independent of the others, then the random variable X= answer "x" questions correctly , has a binomial probability distribution. Then

P(X=x)= n!/((n-x)!*x!)*p^x*(1-p)^(n-x)

where

n= total number of questions= 25

p= probability of getting a question right = 1/4

then

a) P(x=n) = p^n = (1/4)²⁵ = 8*88*10⁻⁶ ( 0.00088 %)

b) P(x<5)= F(5)

where F(x) is the cumulative binomial probability distribution- Then from tables

P(x<5)= F(5)= 0.2137 (21.37%)

7 0
4 years ago
A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature
Westkost [7]

Answer:

(a) The value of <em>k</em> is \frac{1}{15}.

(b) The probability that at most three forms are required is 0.40.

(c) The probability that between two and four forms (inclusive) are required is 0.60.

(d)  P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5 is not the pmf of <em>y</em>.

Step-by-step explanation:

The random variable <em>Y</em> is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of <em>k</em>.

\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}

Thus, the value of <em>k</em> is \frac{1}{15}.

(b)

Compute the value of P (Y ≤ 3) as follows:

P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\   =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5 to be the pmf of Y it has to satisfy the conditions:

  1. P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\
  2. \sum P(y)=1

<u>Check condition 1:</u>

y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0

Condition 1 is fulfilled.

<u>Check condition 2:</u>

\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1

Condition 2 is not satisfied.

Thus, P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5 is not the pmf of <em>y</em>.

7 0
3 years ago
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