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3 years ago

Help ASAP, I’ll give brainliest Show work ty

Mathematics

1 answer:

Liula [17]3 years ago

6 0

Step-by-step explanation:

3(6+15)

=3(21)

=63

5[15-(6+1)]

=5(8)

=40

⅓(15)

⅓(6+15)

=⅓(21)

15/6

=2/1/2 or 2.5

2(6)(15-3)

=2(7)

=36

4(⅓+⅓+⅓)

=4(1)

6/15 - 15/6

=36/90 - 225/90

= -189/90

= -21/10

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If a coin is tossed three times, find probability of getting

Assoli18 [71]

${\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}$

‣ A coin is tossed three times.

${\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}$

‣ The probability of getting,

1) Exactly 3 tails

2) At most 2 heads

3) At least 2 tails

4) Exactly 2 heads

5) Exactly 3 heads

${\large{\textsf{\textbf{\underline{\underline{Using \: Formula :}}}}}}$

$\star \: \tt P(E)= {\underline{\boxed{\sf{\red{ \dfrac{ Favourable \: outcomes }{Total \: outcomes} }}}}}$

${\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}$

★ When three coins are tossed,

then the sample space = {HHH, HHT, THH, TTH, HTH, HTT, THT, TTT}

[here H denotes head and T denotes tail]

⇒Total number of outcomes $\tt [ \: n(s) \: ]$ = 8

1) Exactly 3 tails 

Here

• Favourable outcomes = {HHH} = 1

• Total outcomes = 8

$\therefore \sf Probability_{(exactly \: 3 \: tails)} = \red{ \dfrac{1}{8}}$

2) At most 2 heads

[It means there can be two or one or no heads]

Here

• Favourable outcomes = {HHT, THH, HTH, TTH, HTT, THT, TTT} = 7

• Total outcomes = 8

$\therefore \sf Probability_{(at \: most \: 2 \: heads)} = \green{ \dfrac{7}{8}}$

3) At least 2 tails 

[It means there can be two or more tails]

Here

• Favourable outcomes = {TTH, TTT, HTT, THT} = 4

• Total outcomes = 8

$\longrightarrow \sf Probability_{(at \: least \: 2 \: tails)} = \dfrac{4}{8}$

$\therefore \sf Probability_{(at \: least \: 2 \: tails)} = \orange{\dfrac{1}{2}}$

4) Exactly 2 heads 

Here

• Favourable outcomes = {HTH, THH, HHT } = 3

• Total outcomes = 8

$\therefore \sf Probability_{(exactly \: 2 \: heads)} = \pink{ \dfrac{3}{8}}$

5) Exactly 3 heads

Here

• Favourable outcomes = {HHH} = 1

• Total outcomes = 8

$\therefore \sf Probability_{(exactly \: 3 \: heads)} = \purple{ \dfrac{1}{8}}$

$\rule{280pt}{2pt}$

8 0

2 years ago

It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken if the desir

Ksju [112]

Answer:

We need a sample size of at least 75.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, we find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

The standard deviation is the square root of the variance. So:

$\sigma = \sqrt{484} = 22$

With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is

We need a sample size of at least n, in which n is found when M = 5. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$5 = 1.96*\frac{22}{\sqrt{n}}$

$5\sqrt{n} = 43.12$

$\sqrt{n} = \frac{43.12}{5}$

$\sqrt{n} = 8.624$

$(\sqrt{n})^{2} = (8.624)^{2}$

$n = 74.4$

We need a sample size of at least 75.

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