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VMariaS [17]
3 years ago
5

Simplify

iv%204" id="TexFormula1" title="(5 + 1) ^{2} - (11 + 3 ^{2} ) \div 4" alt="(5 + 1) ^{2} - (11 + 3 ^{2} ) \div 4" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Elodia [21]3 years ago
6 0

Answer:

31

Step-by-step explanation:

Evaluate following the order set out in the acronym PEMDAS

That is parenthesis, exponents, multiplication, division, addition, subtraction

Given

(5 + 1)² - (11 + 3² ) ÷ 4

= 6² - (11 + 9) ÷ 4

= 36 - 20 ÷ 4 ← division before subtraction

= 36 - 5

= 31

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Hese are the values in Paul’s data set.
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Answer: See description below.

The residuals are the differences between the predicted values and the actual values. You will need to make a scatter plot of each difference.

Here are the five points that you will need to plot:
20 - 21 = -1  (1, -1)
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3 years ago
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A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature
Westkost [7]

Answer:

(a) The value of <em>k</em> is \frac{1}{15}.

(b) The probability that at most three forms are required is 0.40.

(c) The probability that between two and four forms (inclusive) are required is 0.60.

(d)  P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5 is not the pmf of <em>y</em>.

Step-by-step explanation:

The random variable <em>Y</em> is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of <em>k</em>.

\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}

Thus, the value of <em>k</em> is \frac{1}{15}.

(b)

Compute the value of P (Y ≤ 3) as follows:

P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\   =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5 to be the pmf of Y it has to satisfy the conditions:

  1. P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\
  2. \sum P(y)=1

<u>Check condition 1:</u>

y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0

Condition 1 is fulfilled.

<u>Check condition 2:</u>

\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1

Condition 2 is not satisfied.

Thus, P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5 is not the pmf of <em>y</em>.

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Two hundred people attended the history museum exhibit on Monday. Twenty-five of the people who attended the exhibit were survey
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Answer:

96

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Surveyed people are 25

Out of the 25, 12 people said yes

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To find the total number of people expected to say yes out of those that attended the history museum;

Fraction of the sampled that said yes × total number of people that attended.

= 12/25 × 200

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