<h3>
Answer: 10.1 cm approximately</h3>
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Explanation:
The double tickmarks show that segments DE and EB are the same length.
The diagram shows that DB = 16 cm long
We'll use these facts to find DE
DE+EB = DB
DE+DE = DB
2*DE = DB
DE = DB/2
DE = 16/2
DE = 8
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Now let's focus on triangle DEC. We just found the horizontal leg is 8 units long. The vertical leg is EC which is unknown for now. We'll call it x. The hypotenuse is CD = 9
Use the pythagorean theorem to find x
a^2+b^2 = c^2
8^2+x^2 = 9^2
64+x^2 = 81
x^2 = 81 - 64
x^2 = 17
x = sqrt(17)
That makes EC to be exactly sqrt(17) units long.
If you follow those same steps for triangle ADE, then you'll find the missing length is AE = 6
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So,
AC = AE+EC
AC = 6 + sqrt(17)
AC = 10.1231056256177
AC = 10.1 cm approximately
Answer:
Probability of picking an even and then even number =
Step-by-step explanation:
Probability of picking a card with even number,
P(even 1) = =
=
=
Followed by picking a card which is even without putting the first card back.
So an even card is to be picked out of remaining 3 cards.
P(even 2) =
=
Now probability of both events = P(even 1) × P(even 2)
=
=
Therefore, probability of both the events will be .
Answer:
x = 8, y =
Step-by-step explanation:
Both 8 and x sides are congruent so they're equal.
Add 64 + 64.
Square root 128
Final answers: x = 8, y =
Answer:
0.2364
Step-by-step explanation:
We will take
Lyme = L
HGE = H
P(L) = 16% = 0.16
P(H) = 10% = 0.10
P(L ∩ H) = 0.10 x p(L U H)
Using the addition theorem
P(L U H) = p(L) + P(H) - P(L ∩ H)
P(L U H) = 0.16 + 0.10 - 0.10 * p(L u H)
P(L U H) = 0.26 - 0.10p(L u H)
We collect like terms
P(L U H) + 0.10P(L U H) = 0.26
This can be rewritten as:
P(L U H)[1 +0.1] = 0.26
Then we have,
1.1p(L U H) = 0.26
We divide through by 1.1
P(L U H) = 0.26/1.1
= 0.2364
Therefore
P(L ∩ H) = 0.10 x 0.2364
The probability of tick also carrying lyme disease
P(L|H) = p(L ∩ H)/P(H)
= 0.1x0.2364/0.1
= 0.2364
Answer:
10
Step-by-step explanation:
yellow
6+6+3=15
so
for green
4+4+2=10