Give the quadratic sequence 2,3,10,23,........ write down the next term of the sequence.

Try the following example. Remember to use the hints if you need them. Solve for x. 2.9×10−7=(x)(x)2(4.6)(4.6)2(3)2 Choose the c

galben [10]

Answer:

$2.3\times 10^{-2}$

Step-by-step explanation:

Given equation,

$2.9\times 10^{-7}=\frac{(x)(x)^2(4.6)}{(4.6)^2(3)^2}$

$2.9\times 10^{-7}=\frac{x^3}{9\times 4.6}$

$(\because a^m = \frac{1}{a^{-m}})$

By cross multiplication,

$2.9\times 9\times 4.6\times 10^{-7}=x^3$

$120.06\times 10^{-7}=x^3$

$\implies x = (120.06\times 10^{-7})^\frac{1}{3}$

$=0.0228980999294$

$\approx 0.023$

$=2.3\times 10^{-2}$

7 0

3 years ago

A number is decreased by 8

astra-53 [7]

Answer:

x - 8

Step-by-step explanation:

I hope this helps.

4 0

3 years ago

Find the slope of the line through each pair of points (12, -18) & (-15, -18)

padilas [110]

Answer:

0

Step-by-step explanation:

Use rise over run, (y2 - y1) / (x2 - x1)

Plug in the points:

(y2 - y1) / (x2 - x1)

(-18 + 18) / (-15 - 12)

0 / -27

= 0

So, the slope is 0.

8 0

3 years ago

How do I find the area AND distance of this irregular shape?

elena-14-01-66 [18.8K]

Find the area of the square first using A=bh then find the semicircle's using A=pi (r) ^2. After finding both, add them. The sum of both will be the total area.

7 0

4 years ago

The age of the children in kindergarten on the first day of school is uniformly distributed between 4.8 and 5.8 years old. A fir

Kazeer [188]

Answer:

(1) (c) 5.30 years.

(2) (b) 0.289.

(3) (b) 0.80.

(4) (d) 0.50.

(5) (a) 5.25 years.

Step-by-step explanation:

Let X = age of the children in kindergarten on the first day of school.

The random variable X follows a continuous Uniform distribution with parameters a = 4.8 years and b = 5.8 years.

The probability density function function of X is:

$f_{X}(x)=\left \{ {{\frac{1}{b-a}} ;\ a$

(1)

The expected value of a Uniform random variable is:

$E(X)=\frac{1}{2}(a+b)$

Compute the mean of X as follows:

$E(X)=\frac{1}{2}(a+b)=\frac{1}{2}\times (4.8+5.8)=5.3$

Thus, the mean of the distribution is (c) 5.30 years.

(2)

The standard deviation of a Uniform random variable is:

$SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}$

Compute the standard deviation of X as follows:

$SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}=\sqrt{\frac{1}{12}\times (5.8-4.8)^{2}}=0.289$

Thus, the standard deviation of the distribution is (b) 0.289.

(3)

Compute the probability that a randomly selected child is older than 5 years old as follows:

$P(X>5)=\int\limits^{5.8}_{5} {\frac{1}{5.8-4.8}}\, dx\\$

$=\int\limits^{5.8}_{5} {1}\, dx\\=[x]^{5.8}_{5}\\=(5.8-5)\\=0.8$

Thus, the probability that a randomly selected child is older than 5 years old is (b) 0.80.

(4)

Compute the probability that a randomly selected child is between 5.2 years and 5.7 years old as follows:

$P(5.2$

$=\int\limits^{5.7}_{5.2} {1}\, dx\\=[x]^{5.7}_{5.2}\\=(5.7-5.2)\\=0.5$

Thus, the probability that a randomly selected child is between 5.2 years and 5.7 years old is (d) 0.50.

(5)

It is provided that a randomly selected child is at the 45th percentile.

This implies that:

P (X < x) = 0.45

Compute the value of x as follows:

$P (X < x) = 0.45$

$\int\limits^{x}_{4.8} {\frac{1}{5.8-4.8}}\, dx=0.45$

$\int\limits^{x}_{4.8} {1}\, dx=0.45$

$[x]^{x}_{4.8}=0.45$

$x-4.8=0.45\\$

$x=0.45+4.8\\x=5.25$

Thus, the age of the child at the 45th percentile is (a) 5.25 years.

6 0

3 years ago

Give the quadratic sequence 2,3,10,23,........write down the next term of the sequence.​

Give the quadratic sequence 2,3,10,23,........write down the next term of the sequence.