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2 years ago

12

What it 40x + 24? I can't figure this out.

1 answer:

Dmitrij [34]2 years ago

7 0

8(5x+3)

If you miltiplied 8 x 5x it will give you 40 x

And 8 x 3 is 24

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A five-digit number starts with a number between 4-9 in the first position, with no restrictions on the remaining 4 digits. a

Maurinko [17]

Answer:

(a) $Pr = 0.3024$

(b) $Pr = 0.6976$

(c) $Pr = \frac{^9P_{n-1}}{10^{n-1}}$

Step-by-step explanation:

Given

$Start = \{5,6,7,8\}$ i.e. between 4 and 9

$n(Start) =4$

$Digits = 5$

Solving (a): Probability that each of the 5 digit are different

Since there is no restriction;

The total possible selection is as follows:

$First\ digit = 4$ (i.e. any of the 4 start digits)

$Second\ digit = 10\\$ (i.e. any of the 10 digits 0 - 9)

$Third\ digit = 10$ (i.e. any of the 10 digits 0 - 9)

$Fourth\ digit = 10$ (i.e. any of the 10 digits 0 - 9)

$Fifth\ digit = 10$ (i.e. any of the 10 digits 0 - 9)

So, the total is:

$Total = 4 * 10 * 10 * 10 * 10$

$Total = 40000$

For selection that all digits are different, the selection is:

$First\ digit = 4$ (i.e. any of the 4 start digits)

$Second\ digit = 9$ (i.e. any of the remaining 9)

$Third\ digit = 8$ (i.e. any of the remaining 8)

$Fourth\ digit = 7$ (i.e. any of the remaining 7)

$Fifth\ digit = 6$ (i.e. any of the remaining 6)

So:

$Selection =4 * 9 * 8 * 7 * 6$

$Selection =12096$

So, the probability is:

$Pr = \frac{Selection}{Total}$

$Pr = \frac{12096}{40000}$

$Pr = 0.3024$

Solving (b): At least 1 repeated digit

The probability calculated in (a) is the all digits are different i.e. P(None)

So, using laws of complement

We have:

$P(At\ least\ 1) = 1 - P(None)$

So, we have:

$Pr= 1 - 0.3024$

$Pr = 0.6976$

Solving (c): An expression to model the probability.

<em>Using (a) as a point of reference, we have;</em>

$Pr = \frac{Selection}{Total}$

Where

$Selection =4 * 9 * 8 * 7 * 6$ ---- for selection of 5 i.e. n = 5

$Total = 4 * 10 * 10 * 10 * 10$

$Selection =4 * 9 * 8 * 7 * 6$

This can be rewritten as:

$Selection = 4 * ^9P_4$

4 can be expressed as: 5 - 1

So, we have:

$Selection = (5-1) *^9P_{5-1}$

Substitute n for 5

$Selection = (n-1) *^9P_{n-1}$

$Selection = (n-1)^9P_{n-1}$

$Total = 4 * 10 * 10 * 10 * 10$

This can be rewritten as:

$Total = 4 * 10^4$

$Total = (5-1) * 10^{5-1}$

$Total = (n-1) * 10^{n-1}$

$Total = (n-1) 10^{n-1}$

So, the expression is:

$Pr = \frac{(n-1)^9P_{n-1}}{(n-1)10^{n-1}}$

$Pr = \frac{^9P_{n-1}}{10^{n-1}}$

<em>Where n represents the digit number</em>

5 0

3 years ago

Can some one please help and please hurry

fenix001 [56]

Hi don’t know but have a wonderful day

8 0

3 years ago

FG ∥ CB, A ∈ FG, D ∈ AB, E ∈ AC

andrey2020 [161]

Answer:

The value of x is $31^{\circ}$ .

Step-by-step explanation:

Please look at the figure attached to get more clear solution.

We have given:

FG||CB

And the line that cut the parallel line is transversal so, here BA is transversal

And alternate interior angles on transverse line are equal

So, ∠1=∠4

And ∠4= $28^{\circ}$

Hence, ∠1=∠4= $28^{\circ}$

And On FG the sum of angles will be $180^{\circ}$

∠3+∠2+∠1= $180^{\circ}$

$90^{\circ}$ +∠2+ $28^{\circ}$ = $180^{\circ}$

Hence, ∠2= $62^{\circ}$

Now, we know that the sum of interior angles is equal to the exterior angle:

Therefore, ∠2+∠5=∠6+∠7

$62^{\circ}+3x=x+4x$

On simplification we get:

$2x=62^{\circ}$

$x=31^{\circ}$

Hence, the value of x is $31^{\circ}$ .

3 0

3 years ago

Shifting a graph vertically and horizontally is only possible for certain types of functions.

dusya [7]

Answer:

B.

Step-by-step explanation:

6 0

3 years ago

I need the answers for this

velikii [3]

Answer:

q1 9.2

q2 5.9

q3 8.4

Q4 7.5

q5 7.3

q6 5.1

5 0

3 years ago