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3 years ago

$240 was shared in the ratio of 2:3:1 by Jacob,

Mathematics

1 answer:

Bingel [31]3 years ago

8 0

Answer:

Jacob: $80

Hilary: $120

Brian: $40

Step-by-step explanation:

Let the amount Brian gets be $x.

Amount Jacob gets= $(2x)

Amount Hilary gets= $(3x)

Total amount of money= $x +$(2x) +$(3x)= $(6x)

$240= $(6x)

6x= 240

x= 240÷6

x= 40

Now substitute the value of x to find out the amount of money each one gets.

Brian gets $40.

Amount Jacob gets= 2(40)= $80

Amount Hilary gets= 3(40)= $120

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What method would be the easiest to solve this equation?

Advocard [28]

Answer:

1, -6

Step-by-step explanation:

Easiest is just to factor

(x+1)(x-6)=0

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3 years ago

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at a particular high school, 42% of the students participate in sports and 25% of the students participate in drama. if 53% of t

baherus [9]

The probability that a student participates in both sports and drama is $0.14$ .

<h3>What is the formula for P(AUB), where A and B are any two events?</h3>

If $A$ and $B$ are any two events, then the probability of the joint event $(A\cup B)$ is given by the following formula: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Given that 42% of the students participate in sports and 25% of the students participate in drama and 53% of the students participate in either sports or drama.

Suppose $S$ denotes that "a student participates in sports" and $D$ denotes that "a student participates in drama".
So, we have $P(S)=\frac{42}{100}=0.42$ , $P(D)=\frac{25}{100}=0.25$ , $P(S\cup D)=\frac{53}{100}=0.53$ .

We want to find the probability that a student participates in both sports and drama i.e., we want to find $P(S\cap D)$ .

By the above formula, we obtain:

$P(S\cup D)=P(S)+P(D)-P(S\cap D)\\\Longrightarrow P(S\cap D)=P(S)+P(D)-P(S\cup D)\\\Longrightarrow P(S\cap D)=0.42+0.25-0.53\\\therefore P(S\cap D)=0.14=\frac{14}{100}$

Therefore, the probability that a student participates in both sports and drama is $0.14$ .

To learn more about probability, refer: brainly.com/question/24756209

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2 years ago

Read 2 more answers

How many students participated in the survey ?

-Dominant- [34]

50 students participated in the survey, if you need help with 2, 3, & 4, just comment and ask

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3 years ago

In a recent survey of college professors, it was found that the average amount of money spent on entertainment each week was nor

Alenkinab [10]

Answer:

0.0918

Step-by-step explanation:

We know that the average amount of money spent on entertainment is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.

The mean and standard deviation of average spending of sample size 25 are

μxbar=μ=95.25

σxbar=σ/√n=27.32/√25=27.32/5=5.464.

So, the average spending of a sample of 25 randomly-selected professors is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.

The z-score associated with average spending $102.5

Z=[Xbar-μxbar]/σxbar

Z=[102.5-95.25]/5.464

Z=7.25/5.464

Z=1.3269=1.33

We have to find P(Xbar>102.5).

P(Xbar>102.5)=P(Z>1.33)

P(Xbar>102.5)=P(0<Z<∞)-P(0<Z<1.33)

P(Xbar>102.5)=0.5-0.4082

P(Xbar>102.5)=0.0918.

Thus, the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.5 is 0.0918.

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3 years ago

Write the slope-intercept form of the equation that passes through the point (-3, 5) and is perpendicular to the line y = 1/5x +

aleksley [76]

Answer:

Step-by-step explanation:

$\text{Let}\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\=========================$

$\text{We have}\ y=\dfrac{1}{5}x+10\to m_1=\dfrac{1}{5}\\\\\text{Therefore}\ m_2=-\dfrac{1}{\frac{1}{5}}=-5.\\\\\text{Put the value of a slope and the coordinates of the point (-3, 5)}\\\text{to the equation}\ y=mx+b:\\\\5=-5(-3)+b\\5=15+b\qquad\text{subtract 15 from both sides}\\-10=b\to b=-10\\\\\text{Finally:}\\\\y=-5x-10$

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3 years ago