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Ivanshal [37]
2 years ago
11

Can you help thank you for my brother due in 14 min please show your work

Mathematics
2 answers:
Sedaia [141]2 years ago
5 0

Answer:

8

Step-by-step explanation:

4 ÷ .5 = 8

AleksAgata [21]2 years ago
4 0
2 divided by 1/4 is 8
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I keep on getting the wrong answer
Anon25 [30]

1) Area = (b*h)/2 = (9*5)/2 = 45/2 = 22.5 (Letter A)

2) Area = b*h = 8*14 = 112 (Letter D)

3) Surface area of a prism is SA=2B+ph (B = area of the base, p = perimeter of the base, h = height)

B = 15 * 5 = 75 cm^2

p = 15 + 5 = 20 cm

SA = 2*75 + 20*7 = 150 + 140 = 290 (G)

4) V = (B*H*L)/2 = (15*7*5)/2  = 525/2 = 262.5 cm^3 (G)

5) V = 9^3 = 81 cm^3 worth of wrapping (A)

6) V = (B*H*L)/2 = (13*6*8)/2 = 312 cube feet (J)

It gave me ADGGAJ. I don't know if this is right, but I atleast tried to do something, right?

5 0
2 years ago
Read 2 more answers
Please identify the common angles and sides or if there are non. Explain all work please!
vova2212 [387]

Answers:

Common angles: ΔBGC and ΔGDC

Common sides: ΔGE and ΔAG

A common angle is a triangle that exists in two or both of the triangles.

Here, the common angles should be ΔBGC and ΔGDC.

These two angles are triangles, and they consist in both triangles.

A common side is when two angles have one vertex in the same area.

Here, we see that the common sides are ΔGE, because they intersect.

Another common side we see here is ΔAG, because they also intersect.

Hope this answer helps you!

8 0
3 years ago
Does the point (-4, 2) lie inside or outside or on the circle x^2 + y^2 = 25?​
d1i1m1o1n [39]

Given equation of the Circle is ,

\sf\implies x^2 + y^2 = 25

And we need to tell that whether the point (-4,2) lies inside or outside the circle. On converting the equation into Standard form and determinimg the centre of the circle as ,

\sf\implies (x-0)^2 +( y-0)^2 = 5 ^2

Here we can say that ,

• Radius = 5 units

• Centre = (0,0)

Finding distance between the two points :-

\sf\implies Distance = \sqrt{ (0+4)^2+(2-0)^2} \\\\\sf\implies Distance = \sqrt{ 16 + 4 } \\\\\sf\implies Distance =\sqrt{20}\\\\\sf\implies\red{ Distance = 4.47 }

Here we can see that the distance of point from centre is less than the radius.

Hence the point lies within the circle .

4 0
3 years ago
Read 2 more answers
What is sum or difference for 2/5 2/3
Kisachek [45]
6/15 + 10/15 = 16/15 = 1 1/15

6/15 - 10/15 = -4/15
5 0
3 years ago
A particular concentration of a chemical found in polluted water has been found to be lethal to 20% of the fish that are exposed
torisob [31]

Answer:

a) P(X=14)=(20C14)(0.8)^{14} (1-0.8)^{20-14}=0.109

b) P(X\geq 10) = 1-P(X \leq 9) = 0.9994

c) P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]=0.589

d) E(X)= np = 20 *0.8 = 16

Var(X) = np(1-p) = 20*0.8*(1-0.8) = 3.2

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=20, p=1-0.2=0.8)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

For this case we want this probability:

P(X=14)

And using the mass function we have this:

P(X=14)=(20C14)(0.8)^{14} (1-0.8)^{20-14}=0.109

Part b

For this case we want this probability:

P(X \geq 10)

And we can find this using the complement rule:

P(X\geq 10) = 1-P(X

P(X=0)=(20C0)(0.8)^{0} (1-0.8)^{20-0}=1.05x10^{-14}

P(X=1)=(20C1)(0.8)^{1} (1-0.8)^{20-1}=8.39x10^{-13}

P(X=2)=(20C2)(0.8)^{2} (1-0.8)^{20-2}=3.19x10^{-11}

P(X=3)=(20C3)(0.8)^{3} (1-0.8)^{20-3}=7.65x10^{-10}

P(X=4)=(20C4)(0.8)^{4} (1-0.8)^{20-4}=1.30x10^{-8}

P(X=5)=(20C5)(0.8)^{5} (1-0.8)^{20-5}=1.66x10^{-7}

P(X=6)=(20C6)(0.8)^{6} (1-0.8)^{20-6}=1.66x10^{-6}

P(X=7)=(20C7)(0.8)^{7} (1-0.8)^{20-7}=1.33x10^{-5}

P(X=8)=(20C8)(0.8)^{8} (1-0.8)^{20-8}=8.65x10^{-5}

P(X=9)=(20C9)(0.8)^{9} (1-0.8)^{20-9}=0.00046

And if we replace we got:

P(X\geq 10) = 1-P(X \leq 9) = 0.9994

Part c

For this case we want this probability:

P(X \leq 16)

And we can use the complement rule like this:

P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]

P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205

P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137

P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.0576

P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.0115

And if we replace we got:

P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]=0.589

Part d

The expected value is given by:

E(X)= np = 20 *0.8 = 16

Var(X) = np(1-p) = 20*0.8*(1-0.8) = 3.2

3 0
3 years ago
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