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zloy xaker [14]
3 years ago
9

Hunter surveyed 200 of the students in his school about their favorite color. 66 students said their favorite color was blue. Wh

at percentage of the surveyed students said their favorite color was blue?
Mathematics
1 answer:
UNO [17]3 years ago
6 0
33% because 66/200 is .33, and convert that to percentage by multiplying by 100
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HELP ME PLEASEEE!! <br><br><br><br><br><br> 48 degrees 96 degrees x =
Roman55 [17]

Answer:

144°

96+48(further away angles)

8 0
2 years ago
Read 2 more answers
If outliers are discarded, then the retirement savings by residents of Econistan is normally distributed with a mean of $100,000
zavuch27 [327]

Answer:

P(X>117000)=P(\frac{X-\mu}{\sigma}>\frac{117000-\mu}{\sigma})=P(Z>\frac{117000-100000}{20000})=P(z>0.85)

And we can find this probability using the complement rule and the normal standard table or excel:

P(z>0.85)=1-P(z

The firgure attached illustrate the problem

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the retirement savings of a population, and for this case we know the distribution for X is given by:

X \sim N(100000,20000)  

Where \mu=100000 and \sigma=20000

We are interested on this probability

P(X>117000)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>117000)=P(\frac{X-\mu}{\sigma}>\frac{117000-\mu}{\sigma})=P(Z>\frac{117000-100000}{20000})=P(z>0.85)

And we can find this probability using the complement rule and the normal standard table or excel:

P(z>0.85)=1-P(z

The firgure attached illustrate the problem

7 0
2 years ago
Rico and his friend eat some lasagna. Rico eats 1/9 of lasagna, and his friend eats 1/18 of the lasagna. How much of the lasagna
Blizzard [7]

Answer:

\frac{5}{6} of the lasagna is left

<u>Step-by-step explanation:</u>

Given:

Amount of lasagna Rico eats = 1/9

Amount of lasagna His friend eats = 1/18

To Find:

The amount  lasagna is left =?

Solution:

Total lasagna ate by both Rico and his friend = Amount of lasagna Rico eats +  Amount of lasagna His friend eats

Total lasagna ate by both Rico and his friend = \frac{1}{9} + \frac{1}{18}

Total lasagna ate by both Rico and his friend = \frac{2}{19} + \frac{1}{18}

Total lasagna ate by both Rico and his friend = \frac{3}{18}

Total lasagna ate by both Rico and his friend = \frac{1}{6}

The amount  lasagna is left = 1 - Total lasagna ate by both Rico and his friend.

The amount  lasagna is left = 1 - \frac{1}{6}

The amount  lasagna is left =\frac{6-1}{6}

The amount  lasagna is left =\frac{5}{6}

4 0
3 years ago
Radioactive Decay:
Vadim26 [7]

The question is incomplete, here is the complete question:

The half-life of a certain radioactive substance is 46 days. There are 12.6 g present initially.

When will there be less than 1 g remaining?

<u>Answer:</u> The time required for a radioactive substance to remain less than 1 gram is 168.27 days.

<u>Step-by-step explanation:</u>

All radioactive decay processes follow first order reaction.

To calculate the rate constant by given half life of the reaction, we use the equation:

k=\frac{0.693}{t_{1/2}}

where,

t_{1/2} = half life period of the reaction = 46 days

k = rate constant = ?

Putting values in above equation, we get:

k=\frac{0.693}{46days}\\\\k=0.01506days^{-1}

The formula used to calculate the time period for a first order reaction follows:

t=\frac{2.303}{k}\log \frac{a}{(a-x)}

where,

k = rate constant = 0.01506days^{-1}

t = time period = ? days

a = initial concentration of the reactant = 12.6 g

a - x = concentration of reactant left after time 't' = 1 g

Putting values in above equation, we get:

t=\frac{2.303}{0.01506days^{-1}}\log \frac{12.6g}{1g}\\\\t=168.27days

Hence, the time required for a radioactive substance to remain less than 1 gram is 168.27 days.

7 0
2 years ago
Solve the system of linear equations using the elimination method x=3y;3x-2y=14
vladimir1956 [14]

Answer:

<h2>x = 6 and y = 2 → (6, 2)</h2>

Step-by-step explanation:

\left\{\begin{array}{ccc}x=3y&\text{subtract}\ 3y\ \text{from both sides}\\3x-2y=14\end{array}\right\\\\\left\{\begin{array}{ccc}x-3y=0&\text{multiply both sides by (-3)}\\3x-2y=14\end{array}\right\\\underline{+\left\{\begin{array}{ccc}-3x+9y=0\\3x-2y=14\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad7y=14\qquad\text{divide both sides by 7}\\.\qquad\qquad y=2\\\\\text{put the value of y to the first equation:}\\x=3(2)=6

7 0
3 years ago
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