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Igoryamba
3 years ago
13

Please help me b4 11:59 7-1-20

Mathematics
1 answer:
nekit [7.7K]3 years ago
5 0

Answer : The length of QT is 19.

Step-by-step explanation :

As we are given:

Length of RT = 11

Length of RS = 8

As we are given that R is the mid-point of line QS. That means,

Length of RS = Length of QR = 8

Now we have to calculate the length of ST.

Length of ST = Length of RT - Length of RS

Length of ST = 11 - 8

Length of ST = 3

Now we have to calculate the length of QT.

Length of QT = Length of QR + Length of RS + Length of ST

Length of QT = 8 + 8 + 3

Length of QT = 19

Therefore, the length of QT is 19.

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a charitable organization in Lanberry is hosting a black tie benefit. Yesterday, the organization sold 55 regular tickets and 49
jenyasd209 [6]

Answer:

  • regular: $67
  • VIP: $122

Step-by-step explanation:

Let r and v represent the costs of a regular and VIP ticket, respectively. The two sales can be represented by ...

  55r +49v = 9663

  79r +84v = 15541

We can subtract 7 times the second equation from 12 times the first to eliminate the v terms.

  12(55r +49v) -7(79r +84v) = 12(9663) -7(15541)

  660r +588v -553r -588v = 115,956 -108,787 . . . . . eliminate parentheses

  107r = 7169 . . . . . . . simplify

  r = 67 . . . . . . . . . . . divide by 107

Using this value in the first equation, we have ...

  55(67) +49v = 9663

  v = 5978/49 . . . . . . . . . subtract 3685, divide by 49

  v = 122

A regular ticket costs $67; a VIP ticket costs $122.

_____

<em>Additional comment</em>

When using "elimination" to solve a system of equations, you're looking for coefficients of the same variable that are related by small factors. Preferably, one coefficient is the same as, or a small multiple of, the other. Here, 55 and 79 (the coefficients of x) are not related by an integer, or a couple of small integers. On the other hand, the y-coefficients 49 and 84 have a common factor of 7, and are in the ratio 7:12, a pair of small numbers. This is why we chose to eliminate the y-variable.

The x-variable could be eliminated using 55 and -79 as multipliers of the equations. This results in larger numbers, and more chance for error. (Errors tend to creep in when computing or copying large numbers.)

Of course, any of several machine methods could be used to solve these equations, including graphing and matrix-solving functions. Here, we tried to honor the requirement to solve by elimination.

7 0
3 years ago
The distance between Earth and Mars is about 4.8732 × 107 miles. What is the standard written form of this distance?
bixtya [17]
I believe the form you mean is

4.8732 x 10^{7}

Well move the decimal point of the 4.8732...7 places to the right.

48,732,000

Now find that answer. 

The answer is A.

Please leave a rating and a thanks!
7 0
3 years ago
Round 3308.89 to the nearest australian dollar
const2013 [10]
Round 3308.89 to the nearest Australian dollar - $4336 AUD .
3 0
3 years ago
Simplify 3x + 2 (9-5) <br><br>please help....
viktelen [127]
3x+8 
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3 0
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Read 2 more answers
Consider the following hypothesis test. H0: μ ≥ 55 Ha: μ &lt; 55 A sample of 36 is used. Identify the p-value and state your con
Ket [755]

Answer:

Step-by-step explanation:

Given that:

H_o: \mu \ge 55 \ \\ \\ H_1 : \mu < 55

(a) For x = 54 and s = 5.3

The test statistics can be computed as:

Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }

Z = \dfrac{54-55}{\dfrac{5.3}{\sqrt{36}} }

Z = \dfrac{-1}{\dfrac{5.3}{6} }

Z = -1.132

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(-1.132,35,1) = 0.1326

Decision: p-value is greater than significance level; do not reject \mathbf{H_o}

Conclusion: \  There  \ is \  insufficient \  evidence  \  to \  conclude \  that \ \mu < 55

b

For x = 53 and s = 4.6

The test statistics can be computed as:

Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }

Z = \dfrac{53-55}{\dfrac{4.6}{\sqrt{36}} }

Z = \dfrac{-2}{\dfrac{4.6}{6} }

Z = -2.6087

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(-2.6087,35,1) =0.0066

Decision: p-value is < significance level; we reject the null hypothesis.

Conclusion: \  There  \ is \  sufficient \  evidence  \  to \  conclude \  that \mu < 55

c)

For x = 56 and s = 5.0

The test statistics can be computed as:

Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }

Z = \dfrac{56-55}{\dfrac{5.0}{\sqrt{36}} }

Z = \dfrac{56-55}{\dfrac{5.0}{\sqrt{36}} }

Z = 1.2

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(1.2,35,1) = 0.88009

Decision: p-value is greater than significance level; do not reject \mathbf{H_o}

Conclusion: \  There  \ is \  insufficient \  evidence  \  to \  conclude \  that \ \mu < 55

6 0
3 years ago
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