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3 years ago

How many possible outcomes are there when when a coin is tossed 3times

Mathematics

1 answer:

kvv77 [185]3 years ago

4 0

Answer:

Step-by-step explanation:

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT

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The figure shows the graphs of the cost and revenue functions for a company that manufactures and sells small radios. More than

Lyrx [107]

ANSWER

The company must produce and sell more than 700 radios to have a profit.

EXPLANATION

The revenue function is

$R(x)=40x$
and the cost function is

$C(x)=5,600+32x$

The company will break even when total revenue is equal to the total cost.

This implies that,

$40x =5,600+32x$

We group similar terms to get,

$40x - 32x = 5600$

This simplifies to,

$8x = 5600$

Divide both sides by 8 to get,

$x = \frac{5600}{8}$

$x = 700$

The company breaks even by producing and selling 700 radios.

The company must produce and sell more than 700 radios to have a profit.

5 0

3 years ago

HELP FAST HURRY FAST PLZ

Leviafan [203]

Answer:

cant rlly read the question

Step-by-step explanation:

4 0

3 years ago

Can someone plz help me with this

Levart [38]

D.
None of the above. Because your expression would be, 2 + 0.75 x 6.
Hope it helps. ;)

7 0

3 years ago

Suppose that the population mean for income is $50,000, while the population standard deviation is 25,000. If we select a random

Fudgin [204]

Answer:

Probability that the sample will have a mean that is greater than $52,000 is 0.0057.

Step-by-step explanation:

We are given that the population mean for income is $50,000, while the population standard deviation is 25,000.

We select a random sample of 1,000 people.

Let $\bar X$ = sample mean

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = $50,000

$\sigma$ = population standard deviation = $25,000

n = sample of people = 1,000

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability that the sample will have a mean that is greater than $52,000 is given by = P( $\bar X$ > $52,000)

P( $\bar X$ > $52,000) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{52,000-50,000}{\frac{25,000}{\sqrt{1,000} } }$ ) = P(Z > 2.53) = 1 - P(Z $\leq$ 2.53)

= 1 - 0.9943 = 0.0057

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.53 in the z table which has an area of 0.9943.

Therefore, probability that the sample will have a mean that is greater than $52,000 is 0.0057.

5 0

3 years ago

Mystery Boxes: Breakout Rooms

ollegr [7]

Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

Upper half

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$