Write an algebraic rule to describe the translation C(5, –4)- ’(–2, 1).

Mathematics

1 answer:

Orlov [11]3 years ago

5 0

A negative minus a negative is always a positive ;)

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whenever Sabrina visits the gym she lifts weights for 8 minutes and runs on the treadmill for 35 minutes. write two equivalent e

Darya [45]

Your answer will be 8+35×5=215 because she works out 43 minutes total a day (8+35=43) and then you have to do that for 5 more more day so you can do 43 × 5 or 43+43+43+43+43 your answer will come out the same

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3 years ago

5 múltiplos de 7 mayores que 140

expeople1 [14]

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2 years ago

An unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6

SIZIF [17.4K]

Answer:

Probability of rolling at least 4 sixes is 0.01696.

Step-by-step explanation:

We are given that an unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6 times.

The above situation can be represented through binomial distribution;

$P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......$

where, n = number trials (samples) taken = 6 trials

r = number of success = at least 4

p = probability of success which in our question is probability of

rolling a “six", i.e; p = 0.20

Let X = Number of sixes on a die

So, X ~ Binom(n = 6, p = 0.20)

Now, Probability of rolling at least 4 sixes is given by = P(X $\geq$ 4)

P(X $\geq$ 4) = P(X = 4) + P(X = 5) + P(X = 6)

= $\binom{6}{4} \times 0.20^{4} \times (1-0.20)^{6-4}+\binom{6}{5} \times 0.20^{5} \times (1-0.20)^{6-5}+\binom{6}{6} \times 0.20^{6} \times (1-0.20)^{6-6}$

= $15 \times 0.20^{4} \times 0.80^{2}+6 \times 0.20^{5} \times 0.80^{1}+1 \times 0.20^{6} \times 0.80^{0}$

= 0.0154 + 0.00154 + 0.000064

= 0.01696

Therefore, probability of rolling at least 4 sixes is 0.01696.

8 0

3 years ago

1. Choose the best descriptor for the graph of f(x)=e^x+5.

Black_prince [1.1K]

Starting from a parent function $f(x)$ , if you add a constant to the whole function, i.e. you perform the transformation

$f(x) \to f(x)+k$

you shift the graph of the parent function by $k$ units, upwards if $k$ is positive, downwards otherwise.

In this case, so, you're shifting, the graph of $e^x$ 5 units upwards. Since the parent function passes through the point $(0,1)$ , because $e^0=1$ , the shifted graph mantains the same $x$ coordinate, but the $y$ coordinate is increased by 5, because of the 5 units upwards shift.