<u>Define x:</u>
Let the first number be x.
1st number = x
2nd number = x + 1
3rd number = x + 2
<u>Construct equation:</u>
x + x + 1 + x + 2 = 4(x + 1)
3x + 3 = 4(x + 1)
Answer: 3x + 3 = 4(x + 1)
Answer: 20 ounce
Step-by-step explanation: 1.29/16=0.080625 1.49/20=0.0745
The best way to solve this problem is with a simple equation: part = percent (in decimal form) x whole
In this problem, we're trying to solve for the <em>whole </em>amount that the camera can hold, so we'll assign that the variable w.
We can use the information they gave us - 96 is the <em>part</em>, and 24% is the percent. We'll convert that to decimal form - 0.24
Then, we just plug these values back into the above equation:
96 = 0.24w
Divide both sides by 0.24 to get w on it's own, and you have your answer!
400 = w
Therefore, the memory card can hold up to 400 pictures.
Note: this equation is <em>super </em>helpful. It's good to have on hand for any problem involving percents, because you can solve for the part, percent, or whole depending on the problem.
Answer:
Test statistic = - 2.29
Pvalue = 0.0110
Step-by-step explanation:
The hypothesis :
H0 : p = 0.3
H1 : p < 0.3
Test statistic :
z=pˆ−p/√p(1−p)/n
pˆ = 0.2
Z = (0.20 - 0.30) / √(0.30(1 - 0.30) / 110
Z = - 0.1 / √0.0019090
Z = - 0.1 / 0.0436931
Z = - 2.29
Test statistic = -2.29
The Pvalue :
P(Z < -2.29) = 0.0110
- the probability that a person has the virus given that they have tested positive is 0.0151.
- the probability that a person does not have the virus given that they have tested negative is 0.9999
P(A) = 1/600 = 0.0017
P(B) = 0.9 * 0.0017 + 0.1 * (1 - 0.0017) = 0.1014
A) P (has the virus | tested positive) = P (tested positive | has the virus) ×
P (has the virus)/ P (tested positive)
= 0.9 × 0.0017/0.1014
= 0.0151
B) P (does not have the virus | tested negative) = P (tested negative | does not have the virus) × P (does not have the virus)/ P (tested negative)
= (1 - 0.1) *× (1 - 0.0017)/ (1 - 0.1014)
= 0.9999
Probability is the department of mathematics regarding numerical descriptions of ways likely an occasion is to occur, or how possibly it's far that a proposition is genuine. The possibility of an occasion varies between zero and 1, wherein, roughly speaking, 0 suggests the impossibility of the occasion and 1 shows certainty. The better the possibility of an event, the more likely it is that the event will arise.
A simple instance is the tossing of an honest (unbiased) coin. since the coin is truthful, the 2 results ("heads" and "tails") are both equally likely; the possibility of "heads" equals the chance of "tails"; and considering the fact that no different results are feasible, the possibility of both "heads" or "tails" is 1/2 (that could additionally be written as 0.5 or 50%).
To learn more about Probability visit here:
brainly.com/question/11234923
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