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Viefleur [7K]
2 years ago
8

How many solutions exist for the given equation?

Mathematics
1 answer:
Soloha48 [4]2 years ago
7 0

The number of solutions which exist for the given equation is; Zero

<h3>Solutions of linear equations</h3>

The given equation is;

  • 3x + 13 = 3(x + 6) + 1

By expansion of the parenthesis;

  • 3x + 13 = 3x + 18 + 1

  • 3x + 13 = 3x +19

Hence, since the Right side of the equation is not equal to the left side of the equation, we can conclude that the given equation has no solution.

Read more on solutions to equations;

brainly.com/question/15523338

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The length of a rectangle is equal to three times it’s widt. If the perimeter is equal to 96 feet, what is the length of the rec
exis [7]

Answer:

12 feet

Step-by-step explanation:

If we use a rectangle. there would be 4 sides the length we can represent as x and the width would result as 3x. If we use the perimeter formula, we would get 8x. 8x would equal 96, the perimeter.

8x= 96

x= 12

The length would be 12

Hope this helped!! :D

4 0
3 years ago
Determine the radius of convergence of the following power series. Then test the endpoints to determine the interval of converge
tresset_1 [31]

Answer:

Radius of the convergence is R = \frac{1}{21}

and,

The Interval of convergence is -\frac{1}{21}

Step-by-step explanation:

Given function : Σ(21x)^k

Now,

Using the ratio test, we have

R = \lim_{n \to \infty} |\frac{21x^{k+1}}{21x^{k}}|

or

R = \lim_{n \to \infty} |\frac{21x^{k}\times21x^{1}}{21x^{k}}|

or

R = \lim_{n \to \infty} |21x^{1}|

now,

for convergence R|x| < 1

Therefore,

\lim_{n \to \infty} |21x^{1}| < 1

or

-\frac{1}{21}

and,

Radius of the convergence is R = \frac{1}{21}

and,

The Interval of convergence is -\frac{1}{21}

8 0
3 years ago
Can someone help me find the evaluation for #20 I need to show my work
givi [52]
(96*4)/8 which is 48
4 0
3 years ago
A manufacturer is interested in the output voltage of a power supply used in a PC. Output voltage is assumed to be normally dist
Luda [366]

The missing values in the question are shown in bold forms below.

A manufacturer is interested in the output voltage of a power supply used in a PC. Output voltage is assumed to be normally distributed, with standard deviation of 0.25 V, and the manufacturer wished to test \mathbf{ H_o : \mu = 10 V  \ against \ H_1 : \mu  \neq 10V}, using  n = 10 units. Statistical Tables and Charts

(a) The critical region is \mathbf{\overline X < 9.83} or \mathbf{\overline X < 10.17} . Find the value of  \mathbf{\alpha }

Answer:

∝ = 0.032   (to 3 decimal place)

Step-by-step explanation:

From the given information:

P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- P \bigg ( \dfrac{X - \mu}{\dfrac{\sigma}{\sqrt{n}}} \bigg )< Z<  \bigg ( \dfrac{X - \mu}{\dfrac{\sigma}{\sqrt{n}}} \bigg ) \bigg )

P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- P \bigg ( \dfrac{9.83 - 10}{\dfrac{0.25}{\sqrt{10}}} \bigg )< Z<  \bigg ( \dfrac{10.17- 10}{\dfrac{0.25}{\sqrt{10}}} \bigg ) \bigg )

P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- P \bigg ( \dfrac{-0.17}{\dfrac{0.25}{\sqrt{10}}} \bigg )< Z<  \bigg ( \dfrac{0.17}{\dfrac{0.25}{\sqrt{10}}} \bigg ) \bigg )

P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- P \bigg (-2.15 \bigg )< Z<  \bigg ( 2.15 \bigg ) \bigg )

From the z - tables;

P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- (0.9842 -0.0158) \bigg )

\alpha = \mathbf{P(\overline X < 9.83 ) + P( X> 10.17) = 0.032}

5 0
3 years ago
A soccer team is having a car wash. The team spent $55 on supplies. They earned $330, including tips. The team's profit is the a
Rus_ich [418]
This problem is very simple just subtract 55 from 330, which is 275. Hope this helped :)
7 0
2 years ago
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