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2 years ago

I need help with this question please?!!!

Mathematics

2 answers:

scoundrel [369]2 years ago

8 0

The answer is -1 because it is subtracting by -2

VashaNatasha [74]2 years ago

6 0

Answer:

it would be 0

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The length of human pregnancies from conception to birth follows a distribution with mean 266 days and standard deviation 15 day

Katena32 [7]

Answer:

a) 81.5%

b) 95%

c) 75%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 266 days

Standard Deviation, σ = 15 days

We are given that the distribution of length of human pregnancies is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(between 236 and 281 days)

$P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{281-266}{15})\\\\= P(-2 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -2)\\= 0.838 - 0.023 = 0.815 = 81.5\%$

b) a) P(last between 236 and 296)

$P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{296-266}{15})\\\\= P(-2 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2)\\= 0.973 - 0.023 = 0.95 = 95\%$

c) If the data is not normally distributed.

Then, according to Chebyshev's theorem, at least $1-\dfrac{1}{k^2}$ data lies within k standard deviation of mean.

For k = 2

$1-\dfrac{1}{(2)^2} = 75\%$

Atleast 75% of data lies within two standard deviation for a non normal data.

Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.

7 0

2 years ago

The area of a triangle that is similar to the one below is the area of this triangle. What is the length of the base of the simi

svp [43]

The answer is 7 feet

7 0

2 years ago

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

(x+3)^2 + 16/(x+3)^2 = 92 <br><br> x + 1/(x+3) = ??

Irina-Kira [14]

One
Remark
This is very complicate unless you pick the right method. It's very handy o know about substitutions.

Method
Let z = (k + 2)^2

Now the problem becomes
z + 16/z = 92 Multiply through by z

Solution
z^2 + 16 = 92z
That does not look very promising. If you know the quadratic formula, this mess can be solved. If you do not know what the quadratic formula is, then what I've written is the answer.

Worse yet, you have to know what complex numbers are. Is this something you know about? The z form of this equation is fine. It gives answers that are rational. The problem is that both are negative and so in your next step, you will be forced to take the square root of a negative number.

Maybe the answer is just
(x + 3)^ + 16 = 92(x + 3)^2
If all you have to do is expand this then you get
x^2 + 6x + 9 + 16 = 92(x^2 + 6x + 9) Remove the brackets.
x^2 + 6x + 25 = 92x^2 + 552x + 828 Put the left over to the right.
0 = 92x^2 - x^2 + 552x - 6x + 828 - 25
0 = 91x^1 + 546x + 803

It looks that way from the second question. If I'm wrong about that, put a comment down below.

Two
Put over a common denominator and expand
$\frac{(x + 3)^2 + 1}{(x + 3)^2} = \frac{x^2 + 6x + 9 + 1}{x^2 + 6x + 9} = \frac{x^2 + 6x + 10}{x^2 + 6x + 9}$

3 0

2 years ago

Mona had some colored papers. The teacher gave her 12 more papers. she counted all the

timama [110]

Answer:

Step-by-step explanation:

really need more information about the subject, but im assuming its just basic addition.

36-12=24

8 0

3 years ago