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In ΔPQR, what is the m∠Q?.

charle [14.2K]

The sum of the 3 inside angles of a triangle needs to equal 180:

11x-5 + 6x + 5 + c = 180

Combine like terms:

18x = 180

Divide both sides by 18:

X = 10

Solve for Q:

11x -5= 11(10) -5 = 110-5 = 105 degrees

7 0

2 years ago

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The following table shows scores obtained in an examination by B.Ed JHS Specialism students. Use the information to answer the q

Makovka662 [10]

Answer:

(a) The cumulative frequency curve for the data is attached below.

(b) (i) The inter-quartile range is 10.08.

(b) (ii) The 70th percentile class scores is 0.

(b) (iii) the probability that a student scored at most 50 on the examination is 0.89.

Step-by-step explanation:

(a)

To make a cumulative frequency curve for the data first convert the class interval into continuous.

The cumulative frequencies are computed by summing the previous frequencies.

The cumulative frequency curve for the data is attached below.

(b)

(i)

The inter-quartile range is the difference between the third and the first quartile.

Compute the values of Q₁ and Q₃ as follows:

Q₁ is at the position:

$\frac{\sum f}{4}=\frac{100}{4}=25$

The class interval is: 34.5 - 39.5.

The formula of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 25 = 34.5

(CF) $_{p}$ = cumulative frequency of the previous class = 24

f = frequency of the class interval = 20

h = width = 39.5 - 34.5 = 5

Then the value of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

$=34.5+[\frac{25-24}{20}]\times5\\\\=34.5+0.25\\=34.75$

The value of first quartile is 34.75.

Q₃ is at the position:

$\frac{3\sum f}{4}=\frac{3\times100}{4}=75$

The class interval is: 44.5 - 49.5.

The formula of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 75 = 44.5

(CF) $_{p}$ = cumulative frequency of the previous class = 74

f = frequency of the class interval = 15

h = width = 49.5 - 44.5 = 5

Then the value of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

$=44.5+[\frac{75-74}{15}]\times5\\\\=44.5+0.33\\=44.83$

The value of third quartile is 44.83.

Then the inter-quartile range is:

$IQR = Q_{3}-Q_{1}$

$=44.83-34.75\\=10.08$

Thus, the inter-quartile range is 10.08.

(ii)

The maximum upper limit of the class intervals is 69.5.

That is the maximum percentile class score is 69.5th percentile.

So, the 70th percentile class scores is 0.

(iii)

Compute the probability that a student scored at most 50 on the examination as follows:

$P(\text{Score At most 50})=\frac{\text{Favorable number of cases}}{\text{Total number of cases}}$

$=\frac{10+4+10+20+30+15}{100}\\\\=\frac{89}{100}\\\\=0.89$

Thus, the probability that a student scored at most 50 on the examination is 0.89.

5 0

3 years ago

Evaluate the expression (4x^3)^2 for x=2 A.128 B.512 C.256 D. 1024

julsineya [31]

Answer:

d:) 1024

Step-by-step explanation:

Evaluate (4 x^3)^2 where x = 2:

(4 x^3)^2 = (4×2^3)^2

Multiply each exponent in 4×2^3 by 2:

4^2 (2^3)^2

Multiply exponents. (2^3)^2 = 2^(3×2):

2^(3×2)×4^2

4^2 = 16:

2^(3×2)×16

3×2 = 6:

2^6×16

2^6 = (2^3)^2 = (2×2^2)^2:

(2×2^2)^2 16

2^2 = 4:

(2×4)^2 16

2×4 = 8:

8^2×16

8^2 = 64:

64×16

| | 6 | 4

× | | 1 | 6

| 3 | 8 | 4

| 6 | 4 | 0

1 | 0 | 2 | 4:

Answer: 1024

8 0

2 years ago

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A patio was to be laid in a design with one tile in the

Vlada [557]

Answer:

10

Step-by-step explanation:

The number of tiles in the design is 1 + 2 + 3 + ...

We can model this as an arithmetic series, where the first term is 1 and the common difference is 1. The sum of the first n terms of an arithmetic series is:

S = n/2 (2a₁ + d (n − 1))

Given that a₁ = 1 and d = 1:

S = n/2 (2(1) + n − 1)

S = n/2 (n + 1)

Since S ≤ 60:

n/2 (n + 1) ≤ 60

n (n + 1) ≤ 120

n must be an integer, so from trial and error:

n ≤ 10

Mr. Tong should use 10 tiles in the final row to use the most tiles possible.

7 0

2 years ago

The mechanics at Giuseppe’s Auto House specialize in changing fuel injection units and transmissions. Last week, they changed 5

Lemur [1.5K]

Answer:

The solution is (2,6)

Step-by-step explanation:

Let

x-----> the number of hours to change a fuel injection unit

y-----> the number of hours to change a transmission

we know that

5x+10y=70 -----> equation A

8x+8y=64 -----> equation B

Solve the system of equations by graphing

Remember that the solution of the system of equations is the intersection point both graphs

The solution is the point (2,6)

see the attached figure

4 0

2 years ago

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