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MakcuM [25]
3 years ago
11

5y+4x=10.51 8y+3x=10.39

Mathematics
1 answer:
Phoenix [80]3 years ago
7 0

Answer:

  (x, y) = (1.89, 0.59)

Step-by-step explanation:

Written in general form, the equations are ...

  4x +5y -10.51 = 0

  3x +8y -10.39 = 0

These can be solved using the cross-multiplication method to find ...

  ∆1 = (4)(8) -(3)(5) = 17

  ∆2 = (5)(-10.39) -(8)(-10.51) = 32.13

  ∆3 = (-10.51)(3) -(-10.39)(4) = 10.03

Then the values of the variables are ...

  x = ∆2/∆1 = 32.13/17 = 1.89

  y = ∆3/∆1 = 10.03/17 = 0.59

The solution is (x, y) = (1.89, 0.59).

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Which of the following algebraic equations could represent the sentence, "The quotient of a number and 9 is three"? x - 9 = 3 9
Sergio [31]

\frac{x}{9} =3

The quotient(division) of a number(x) and 9 is(equals) 3

[x represents the unknown number]

Your answer is the 4th option

8 0
3 years ago
Solve the following multiplication and division problems.
algol13

Answer:

a. 22.5 km

b. 267.52 m

c. 95 m

d. (4.2)(10^{-5}) km

Step-by-step explanation:

a. You can convert from hectometers to kilometers, and then you can make the multiplication:

\frac{5hm}{10}=0.5km

(3km)(0.5km)(15)=22.5km

b. You can convert from centimeters to meters, and then you can make the multiplication:

\frac{44cm}{100}=0.44m\\(19m)(0.44m)(32)=267.52m

c. You only need to divide 1,140 meters by 12:

\frac{1,140m}{12})=95m

d. You can convert from hectometers to kilometers, from decameters to kilometers and from meters to kilometers, and divide this by 15:

\frac{6hm}{10}=0.6km

\frac{7dam}{100}=0.07km

\frac{5m}{1000}=0.005km

\frac{(3km)(0.6km)(0.07km)(0.005km)}{15}=(4.2)(10^{-5}) km

5 0
4 years ago
How many miles does it take to make the same amount as one third
grin007 [14]
3/6ths according to my math
3 0
4 years ago
A city councilwoman is concerned that the new contractor she hired is taking too long to replace defective streetlights. She wou
satela [25.4K]

Answer:

There is significant evidence to conclude that the replacement time for streetlights under the new contractor is longer than the replacement time under the previous contractor.

Step-by-step explanation:

Given the data :

6.2 7.1 5.4 5.5 7.5 2.6 4.3 2.9 3.7 0.7 5.6 1.7

The hypothesis:

H0: μ = 3.2

H1 : μ > 3.2

n = sample size = 12

The sample mean, xbar = ΣX / n

xbar = 53.2 / 12

xbar = 4.43

Using calculator;

Sample standard deviation, s = 2.147

The test statistic :

(xbar - μ) ÷ (s/sqrt(n))

(4.43 - 3.2) ÷ (2.147/sqrt(12)

1.23 / 0.6197855

Test statistic = 1.985

The Pvalue using the Pvalue from Tscore calculator :

Tscore = 1.985 ; df = 12 - 1 = 11

Pvalue = 0.036

Since Pvalue < α ; We reject the Null

Hence, we conclude that the replacement time for streetlights under the new contractor is longer than the replacement time under the previous contractor.

5 0
3 years ago
Kenta said “The product of two fractions can sometimes be less than the quotient of the two numbers and sometimes greater.”
Nikolay [14]
The Answer Is True .
4 0
3 years ago
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