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elena55 [62]
3 years ago
8

Twenty is no greater than the sum of a number and -4.

Mathematics
1 answer:
Aleksandr [31]3 years ago
3 0

Hey there!

The statement above would be written like so...

20 < n + -4

Hope this helps you!

God bless ❤️

Happy Halloween!

xXxGolferGirlxXx

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WHAT IS THE SOLUTION TO 5-2N= 17
ryzh [129]
N is equal to 11 because 5-2n=17 would be changed into 2n-5=17 because it would be -17 if it was the original problem so that is why 

7 0
3 years ago
Read 2 more answers
5. Naomi was watching a game show where
Solnce55 [7]

Answer:

C) 50

Step-by-step explanation:

This is because -10 x 5 = -50 and if you have 20 points to start with, you just do 20-(-30)=20 and two minuses make a plus so it is 20 + 30 which is 50.

8 0
3 years ago
A line has a slope of -3/4 and has a y intercept of 3. What is the x intercept of the line
natima [27]

Answer:

The x-intercept is 4

Step-by-step explanation:

We are given;

The slope of a line as -3/4

y-intercept is 3

We are supposed to determine the x-intercept;

We need to know that;

The slope of a line is given by the ratio of the change in y to the change in x

The coordinates of y-intercept are (0, 3)

Taking the x-intercept as a, then the coordinates of x-intercept is (a, 0)

But;

slope = Δ in y ÷ Δ in x

Therefore;

\frac{0-3}{a-0}= \frac{-3}{4}

Solving for a

-3a = -12 \\    a =4

Therefore, the x-intercept is 4

3 0
3 years ago
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}.

4 0
3 years ago
Jessica calculated the missing side length of one of these triangles using the Pythagorean Theorem. Which triangle was it?
Alenkinab [10]

Answer:

G

Step-by-step explanation:

We can find a missing length of a triangle using a Pythagorean theorem if and only the triangle is a right angled triangle.

The side of the missing length is:

a^+b^=c^

2^+4^=c^

4+16=c^

20=c^

\sqrt{20}  = c ^{2}  \\ 4 \sqrt{5} = c

8 0
3 years ago
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