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2 years ago

WILL MARK BRAINLIEST, JUST HELP ME PLS

Mathematics

1 answer:

Luda [366]2 years ago

5 0

Answer:

SAS theorem

Step-by-step explanation:

If $\frac{CE}{CB} = \frac{DC}{AC}$ and m∠ECD = m∠BCA, then the triangles are similar by the SAS theorem.

$\frac{EC}{BC} = \frac{14}{21} = \frac{2}{3}$ $\frac{DC}{AC} = \frac{18}{27} = \frac{2}{3}$

So, $\frac{CE}{CB} = \frac{DC}{AC}$ and m∠ECD = m∠BCA because vertical angles are congruent

Therefore ΔECD is similar to ΔBCA by SAS theorem

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Find the value of x. If your answer is not an integer, express it in simplest radical form.

DedPeter [7]

Answer:

x = 8√5

Step-by-step explanation:

for right triangle

c^2 = a^2 + b^2

with c is the hypotenuse; a & b are the other sides

(4 + 16)^2 = (4√5)^2 + x^2

400 = 80 + x^2

x^2 = 320

x^2 = 64(5)

√x^2 = √64(√5)

x = 8√5

7 0

2 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

(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

Sylvia ran 3km 290m in the morning then she ran some more in the evening if she ran a total of 10km how far did Sylvia run in th

MAXImum [283]

Answer:

She ran 6.71km in the evening

Step-by-step explanation:

3km 290m = 3.29km

10 - 3.29 = 6.71km

4 0

3 years ago

Please help me on this because I'm not sure if it's a decimal or not

anzhelika [568]

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3 0

3 years ago

A conjecture and the paragraph proof used to prove the conjecture are shown.

arlik [135]

Answer:

1. DF (option 3)

2. segment addition postulate

3. segment congruence postulate

4. DF (option 5)

6 0

3 years ago

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