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2 years ago

13

A triangle has side lengths of 33,56, and 65. Is it a right triangle? Explain.

2 answers:

Mariulka [41]2 years ago

7 0

You would've have to put 33, 56, and 65 in the Pythagorean thrum. So, 33^2 + 56^2 = 65^2,
Then solve. 1089 + 3136 = 4225, 4225 = 4225. So A is the correct answer

soldier1979 [14.2K]2 years ago

5 0

U can see if it is a right triangle bu using the pythagorean theorem (or however it is spelled)

a^2 + b^2 = c^2....where a and b are the legs and c is the hypotenuse (the largest side)

so lets sub our numbers in and see if they equal
33^2 + 56^2 = 65^2
1089 + 3136 = 4225
4225 = 4225.......correct...so YES, this is a right triangle

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Imagine slicing the pyramid with a plane parallel to its base and passing through the topmost point of the pyramid. What will th

Nitella [24]

Answer:

Cross sections perpendicular to the base and through the vertex will be triangles. Below, you can see a plane cutting through the pyramid, part of the pyramid removed, and the cross section. You could also take a slice parallel to the base. Cross sections parallel to the base will be hexagons.

Step-by-step explanation:

6 0

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.

5 0

3 years ago

Marla’s School has 1,030 students. Kim’s school has 3 times as many students as Sandi’s School. Write an equation to find the nu

DENIUS [597]

Answer:

3090

Step-by-step explanation:

times 1,030 to 3.

6 0

2 years ago

40.00 jeans 20% discount

Svetllana [295]

20(40)
$\frac{20}{100}$ = 0.2
0.2(40) = 8

After price: $32.00
But, you save: $8

4 0

2 years ago

Kim rode her bicycle 135 miles in 9 weeks, riding the same distance each week. Eric rode his bicycle 102 miles in 6 weeks, ridin

Juli2301 [7.4K]

Answer:

Eric rode 2 more miles per week than Kim rode.

Step-by-step explanation:

Find how many miles per week Kim rode by dividing the number of miles she rode by the number of weeks:

135/9

= 15

Find how many miles per week Eric rode:

102/6

= 17

So, Kim rode 15 miles per week and Eric rode 17 miles per week.

The answer is Eric rode 2 more miles per week than Kim rode.

4 0

3 years ago