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aleksley [76]
3 years ago
13

Someone help plz idk this i have 2 more questions like this

Mathematics
1 answer:
patriot [66]3 years ago
4 0

Answer:

BABE WHERE ARE YOU AT!?

Step-by-step explanation:

like i been waiting for hours now

if u wanted some space TELL ME, dont ghost me.

like im over here lowkey freaking out where are u???????

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What is the slope of the line?
Xelga [282]

Answer:

3/4

Step-by-step explanation:

7 0
3 years ago
On Texas Avenue between University Drive and George Bush Drive, accidents occur according to a Poisson process at a rate of thre
Zarrin [17]

Answer:

(a) The probability is 0.6514

(b) The probability is 0.7769

Step-by-step explanation:

If the number of accidents occur according to a poisson process, the probability that x accidents occurs on a given day is:

P(x)=\frac{e^{-at}*(at)^{x} }{x!}

Where a is the mean number of accidents per day and t is the number of days.

So, for part (a), a is equal to 3/7 and t is equal to 1 day, because there is a rate of 3 accidents every 7 days.

Then, the probability that a given day has no accidents is calculated as:

P(x)=\frac{e^{-3/7}*(3/7)^{x}}{x!}

P(0)=\frac{e^{-3/7}*(3/7)^{0}}{0!}=0.6514

On the other hand the probability that February has at least one accident with a personal injury is calculated as:

P(x≥1)=1 - P(0)

Where P(0) is calculated as:

P(x)=\frac{e^{-at}*(at)^{x} }{x!}

Where a is equivalent to (3/7)(1/8) because that is the mean number of accidents with personal injury per day, and t is equal to 28 because 4 weeks has 28 days, so:

P(x)=\frac{e^{-(3/7)(1/8)(28)}*((3/7)(1/8)(28))^{x}}{x!}

P(0)=\frac{e^{-(3/7)(1/8)(28)}*((3/7)(1/8)(28))^{0}}{0!}=0.2231

Finally, P(x≥1) is:

P(x≥1) = 1 - 0.2231 = 0.7769

3 0
4 years ago
2 4 8 16 what is the rule and the sixth number in the sequence answers
aleksklad [387]
The sequence is incrementing by
{2}^{n}
Therefore, the sixth number is
{2}^{6}
Which is 64.
7 0
4 years ago
A circle has centre (3,0) and radius 5. The line y = 2x + k intersects the circle in two points. Find the set
lara [203]

A circle with center (3,0) and radius 5 has equation

(x-3)^2+y^2=25 \iff x^2 + y^2- 6 x  = 16

If we substitute y=2x+k in this equation, we have

x^2+(2x+k)^2-6x=16 \iff 5x^2+(4k-6)x+k^2-16=0

This equation has two solutions (i.e. the line intersects the circle in two points) if and only if the determinant is greater than zero:

\Delta=b^2-4ac=(4k-6)^2-4\cdot 5\cdot (k^2-16)>0

The expression simplifies to

-4 (k^2 + 12 k - 89)>0 \iff k^2 + 12 k - 89

The solutions to the associated equation are

k^2 + 12 k - 89=0 \iff k=-6\pm 5\sqrt{5}

So, the parabola is negative between the two solutions:

-6-5\sqrt{5}

8 0
3 years ago
X +2 times -2+107 = 180
Kryger [21]

Answer:

X = -3/7

Step-by-step explanation:

7 0
3 years ago
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