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3 years ago

Hello good morning!! Anyone watch anime? My favorite is yu yu hakusho or one piece

Mathematics

2 answers:

sasho [114]3 years ago

7 0

Answer:

I like mha one piece bleach naruto and sometimes romance animes

topjm [15]3 years ago

3 0

Answer:

Good Morning my favorite is Attack on Titan.

Step-by-step explanation:

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5 less than a number is equivalent to 1 more than three times the number. the number is

lutik1710 [3]

Answer:

x = -1

Step-by-step explanation:

5 less than a number is equivalent to 1 more than three times the number

number = x

x - 5 = 3x + 1

now im going to get the numbers and variable on different sides

x - 5 + 5 = x

3x + 1 - 3x = 1

x + 3x = 4x

1 - 5 = -4

4x = -4

lastly im going to divide each side by 4

4x / 4 = x

-4 / 4 = -1

x = -1

3 0

3 years ago

Find axis of symetry y=2x^2+4x-10

Tema [17]

The axis of symmetry is x=-1

6 0

3 years ago

Can you please help me with this there is a pic on the top thx

Svetach [21]

It would be 1000 x 6
so the answer would be 6000.
hope this helped but plz help me out and put this answer as brainliest plz

6 0

3 years ago

Read 2 more answers

Sin^2(x/2)=sin^2x Find the value of x

Veronika [31]

$\sin^2\dfrac x2=\sin^2x$
$\dfrac{1-\cos x}2=1-\cos^2x$
$1-\cos x=2-2\cos^2x$
$2\cos^2x-\cos x-1=0$
$(2\cos x+1)(\cos x-1)=0$
$\implies\begin{cases}2\cos x+1=0\\\cos x-1=0\end{cases}$ $\implies\begin{cases}\cos x=-\frac12\\\cos x=1\end{cases}$

The first case occurs in $0\le x$ for $x=\dfrac{2\pi}3$ and $x=\dfrac{4\pi}3$ . Extending the domain to account for all real $x$ , we have this happening for $x=\dfrac{2\pi}3+2n\pi$ and $\dfrac{4\pi}3+2n\pi$ , where $n\in\mathbb Z$ .

The second case occurs in $0\le x$ when $x=0$ , and extending to all reals we have $x=2n\pi$ for $n\in\mathbb Z$ , i.e. any even multiple of $\pi$ .

6 0

3 years ago

A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective

Inessa05 [86]

Answer:

(a) P(X $\leq$ 20) = 0.9319

(b) Expected number of defective light bulbs = 15

Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

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

where, n = number of samples taken = 150

r = number of success

p = probability of success which in our question is % of bulbs that

are defective, i.e. 10%

Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.

So, Let X = No. of defective bulbs in a box

Mean of X, $\mu$ = $n \times p$ = $150 \times 0.10$ = 15

Standard deviation of X, $\sigma$ = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 3.7

So, X ~ N( $\mu = 15, \sigma^{2} = 3.7^{2})$

Now, the z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X $\leq$ 20) = P(X < 20.5) ---- using continuity correction

P(X < 20.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{20.5-15}{3.7}$ ) = P(Z < 1.49) = 0.9319

(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = $n \times p$ = $150 \times 0.10$ = 15.

5 0

3 years ago