All categories

3 years ago

Only t circled ones somebody plz help I am gonna get a 0 plz help I will give you brainleist

Mathematics

1 answer:

Vikki [24]3 years ago

4 0

Step-by-step explanation:

1) A number m is at least four

m ≥ 4 ['at least' = ≥ symbol]

2) 4x ≤ 20x; x=2

4(2) ≤ 20(2)

8 ≤ 40

[Im not sure if this is a solution or and inequality :/ ]

3) x < 5

--> On the line, draw the number 0, and the number five. 'x < 5' also means 'x is less that five.' So, draw a dot over the line before 5, and make an arrow going to the left of the page,

<-----------------------------------o 5

<---|---|---|---|---|---|---|---|---|---|---|---|---|---|--->

Hopefully this helps! :)

You might be interested in

Can someone please help me with this:)

madam [21]

Answer: c

Step-by-step explanation:

3 0

3 years ago

За +6+5а Can you write an equivalent expression by combining like terms.

Alexxx [7]

Answer:

It's 8a + 6

Plz mark brainiest thx

4 0

2 years ago

erma4kov [3.2K]

1/4x3/5-1/12x+1/10.

5 0

2 years ago

What is the slope of the line through (-1,2) and (-3,-2)?

Morgarella [4.7K]

Answer:

Step-by-step explanation:

4 0

3 years ago

Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving

lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

$P(x=0) = ^nC_x P^x (1-P)^n^-^x$

$P(x=0) = ^5C_0 (0.4)^0 (1-0.4)^5^-^0$

$P(x=0) = ^5C_0 (0.4)^0 (0.60)^5$

$P(x=0) = 0.778$

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

$= 1 - P(X = 0)$

$= 1 - ^5C_0 (0.4)^0 * (0.6)^5$

$= 1 - 0.0778$

$= 0.9222$

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

$^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 (0.4)^2 (0.6)^3$

$= 0.6826$

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

$= 1 - [^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 * (0.4)^2 (0.6)^3]$

$= 1 - 0.6826$

$= 0.3174$

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

7 0

3 years ago