Please answer asap! If 2 or more people answer I will give brainliest

What is the theoretical probability of the spinner landing on a 5

Katarina [22]

The correct answer is 1/8.

6 0

3 years ago

5 and below is having a discount sale for 8% off on all items. She wanted to get a shirt that was originally $25.00. How much wi

Lena [83]

AnswerNaruto fan?

Step-by-step explanation:

3 0

4 years ago

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Find area of trapezoids

ElenaW [278]

Answer:

Area of trapezoid = 660 sq.in

Step-by-step explanation:

here's the solution : -

Area of trapezoid : -

=》

$\frac{1}{2} \times h(b1 + b2)$

=》

$\frac{1}{2} \times 20 \times (30 + 36)$

=》

$10 \times 66$

=》

$660$

8 0

3 years ago

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A computer system uses passwords that are exactly six characters and each character is one of the 26 letters (a–z) or 10 integer

ELEN [110]

Answer:

The number of hits would follow a binomial distribution with $n =10,\!000$ and $p \approx 4.59 \times 10^{-6}$ .

The probability of finding $0$ hits is approximately $0.955$ (or equivalently, approximately $95.5\%$ .)

The mean of the number of hits is approximately $0.0459$ . The variance of the number of hits is approximately $0.0459\!$ (not the same number as the mean.)

Step-by-step explanation:

There are $(26 + 10)^{6} \approx 2.18 \times 10^{9}$ possible passwords in this set. (Approximately two billion possible passwords.)

Each one of the $10^{9}$ randomly-selected passwords would have an approximately $\displaystyle \frac{10,\!000}{2.18 \times 10^{9}}$ chance of matching one of the users' password.

Denote that probability as $p$ :

$p := \displaystyle \frac{10,\!000}{2.18 \times 10^{9}} \approx 4.59 \times 10^{-6}$ .

For any one of the $10^{9}$ randomly-selected passwords, let $1$ denote a hit and $0$ denote no hits. Using that notation, whether a selected password hits would follow a bernoulli distribution with $p \approx 4.59 \times 10^{-6}$ as the likelihood of success.

Sum these $0$ 's and $1$ 's over the set of the $10^{9}$ randomly-selected passwords, and the result would represent the total number of hits.

Assume that these $10^{9}$ randomly-selected passwords are sampled independently with repetition. Whether each selected password hits would be independent from one another.

Hence, the total number of hits would follow a binomial distribution with $n = 10^{9}$ trials (a billion trials) and $p \approx 4.59 \times 10^{-6}$ as the chance of success on any given trial.

The probability of getting no hit would be:

$(1 - p)^{n} \approx 7 \times 10^{-1996} \approx 0$ .

(Since $(1 - p)$ is between $0$ and $1$ , the value of $(1 - p)^{n}$ would approach $0\!$ as the value of $n$ approaches infinity.)

The mean of this binomial distribution would be: $n\cdot p \approx (10^{9}) \times (4.59 \times 10^{-6}) \approx 0.0459$ .

The variance of this binomial distribution would be:

$\begin{aligned}& n \cdot p \cdot (1 - p)\\ & \approx(10^{9}) \times (4.59 \times 10^{-6}) \times (1- 4.59 \times 10^{-6})\\ &\approx 4.59 \times 10^{-6}\end{aligned}$ .