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3 years ago

5

Find the surface area

1 answer:

Brut [27]3 years ago

7 0

Answer:

Step-by-step explanation:

area of top=7×6=42 yd²

area of bottom=3×6=18 yd²

area of two rectangles=2[4×6]=48 yd²

area of two trapezoids=2[1/2(7+3)×3.5]=35 yd²

total surface area=42+18+48+35=143 yd²

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Help me more 10 points

nignag [31]

well you have to turn it into a percentage so he got a touch down more than half of his games because 9 is more than half of 12 so if it's more than 50% what is it

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3 years ago

In which quadrant will the triangle be located after this series of transformations? Does the size of the triangle change after

ASHA 777 [7]

Answer:

Step-by-step explanation:

1. Translate (x, y) → (x + 1, y + 2)

The x coordinate will be translated 1 units horizontally and the y coordinate will be translated up 2 units vertically but they are still locate in the 1st quadrant

2. R (origin, 90° clockwise)

It still locate in the 1st quadrant because after rotating it will create an angle 90°

3. Translate (x, y) → (x + 2, y - 3)

The x coordinate will be translated 2 units horizontally but the the y coordinate will be translated down 3 units vertically, so:

- It will remain in the first quadrant.

- There is no size of the triangle will change.

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3 years ago

Answer ASAP I am being timed

maria [59]

Answer:

i think its c

Step-by-step explanation:

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3 years ago

Twenty percent of drivers driving between 10 pm and 3 am are drunken drivers. In a random sample of 12 drivers driving between 1

Lesechka [4]

Answer:

(a) 0.28347

(b) 0.36909

(c) 0.0039

(d) 0.9806

Step-by-step explanation:

Given information:

n=12

p = 20% = 0.2

q = 1-p = 1-0.2 = 0.8

Binomial formula:

$P(x=r)=^nC_rp^rq^{n-r}$

(a) Exactly two will be drunken drivers.

$P(x=2)=^{12}C_{2}(0.2)^{2}(0.8)^{12-2}$

$P(x=2)=66(0.2)^{2}(0.8)^{10}$

$P(x=2)=\approx 0.28347$

Therefore, the probability that exactly two will be drunken drivers is 0.28347.

(b)Three or four will be drunken drivers.

$P(x=3\text{ or }x=4)=P(x=3)\cup P(x=4)$

$P(x=3\text{ or }x=4)=P(x=3)+P(x=4)$

Using binomial we get

$P(x=3\text{ or }x=4)=^{12}C_{3}(0.2)^{3}(0.8)^{12-3}+^{12}C_{4}(0.2)^{4}(0.8)^{12-4}$

$P(x=3\text{ or }x=4)=0.236223+0.132876$

$P(x=3\text{ or }x=4)\approx 0.369099$

Therefore, the probability that three or four will be drunken drivers is 0.3691.

(c)

At least 7 will be drunken drivers.

$P(x\geq 7)=1-P(x$

$P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)]$

$P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155]$

$P(x\leq 7)=1-[0.9961]$

$P(x\leq 7)=0.0039$

Therefore, the probability of at least 7 will be drunken drivers is 0.0039.

(d) At most 5 will be drunken drivers.

$P(x\leq 5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)$

$P(x\leq 5)=0.06872+0.20616+0.28347+0.23622+0.13288+0.05315$

$P(x\leq 5)=0.9806$

Therefore, the probability of at most 5 will be drunken drivers is 0.9806.

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3 years ago

Number 4 please asap!!!!!!!

Annette [7]

I would say it's H because 4 and 3 add to be 7

8 0

4 years ago