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Lorico [155]
3 years ago
11

An international calling plan charges a rate per minute plus a flat fee. A 10-minute call to France cost $3.19. A 15-minute call

to France costs $4.29. Write and solve a linear equation to find the cost of a 12-minute call to France
Mathematics
1 answer:
viva [34]3 years ago
6 0

Answer:

$3.63

Step-by-step explanation:

A 10-minute call to France cost $3.19.

(x_1,y_1)=(10,3.19)

A 15-minute call to France costs $4.29.

(x_2,y_2)=(15,4.29)

Now we will use two point slope form

y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)

Substitute the values :

y-3.19=\frac{4.29-3.19}{15-10}(x-10)

y-3.19=0.22(x-10)

y-3.19=0.22x-2.2

y=0.22x-2.2+3.19

y=0.22x+0.99

y is the cost and x is the number of minutes

Now we are suppose to find cost of a 12-minute call to France

Substitute x = 12

y=0.22(12)+0.99

y=3.63

Hence the cost of a 12-minute call to France is $3.63

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OverLord2011 [107]

Answer:

x^(5/6) + 4(x^(7/3))


Step-by-step explanation:

Simplify x to the 1/3 power MULTIPLIED BY (x to the 1/2 power + 2x to the 2 power )


Simplify x^(1/3) × (x^(1/2) + (2x)^2)

= x^(1/3)(x^(1/2)) + x^(1/3)((2x)^2)

= x^(1/3+1/2) + 4(x^(1/3+2))

= x^(5/6) + 4(x^(7/3))


x^(1/3) is y such that y^3 = x

(x^(1/3) × x^(1/3) × x^(1/3)) = x^(1/3+1/3+1/3) = x^1 = x


x^(1/2) = √2 = y such that y^2 = x

(2x)^2 = 4x^2






6 0
3 years ago
X+5+4-5=x+54<br>Help !!!​
max2010maxim [7]
  • Simplify both sides of the equation

=>x+5+4−5=x+54

=>x+5+4+−5=x+54

=>(x)+(5+4+−5)=x+54(Combine Like Terms)

=>x+4=x+54

=>x+4=x+54

  • Subtract x from both sides

=>x+4−x=x+54−x

=>4=54

  • Subtract 4 from both sides

=>4−4=54−4

=>0=50

  • There are no solutions
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Step-by-step explanation:

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Suppose a consumer product researcher wanted to find out whether a highlighter lasted less than the manufacturer's claim that th
11111nata11111 [884]

Answer:

The null hypothesis is H_0: \mu = 14

The alternate hypothesis is H_a: \mu < 14

The test statistic is t = -1.95.

The p-value is of 0.0292. This means that for a level of significance of 0.0292 and higher, there is sufficient evidence to conclude that the  highlighters wrote for less than 14 continuous hours.

Step-by-step explanation:

Suppose a consumer product researcher wanted to find out whether a highlighter lasted less than the manufacturer's claim that their highlighters could write continuously for 14 hours.

At the null hypothesis, we test if the mean is 14 hours, that is:

H_0: \mu = 14

At the alternate hypothesis, we test if the mean is less than 14 hours, that is:

H_a: \mu < 14

The test statistic is:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.

14 is tested at the null hypothesis:

This means that \mu = 14

X = 13.6 hours, s = 1.3 hours. Sample of 40:

In addition to the values of X and s given, we have that n = 40

Test statistic:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{13.6 - 14}{\frac{1.3}{\sqrt{40}}}

t = -1.95

The test statistic is t = -1.95.

P-value:

The p-value of the test is the probability of finding a sample mean lower than 13.6, which is a left tailed test, with t = -1.95 and 40 - 1 = 39 degrees of freedom.

Using a calculator, the p-value is of 0.0292. This means that for a level of significance of 0.0292 and higher, there is sufficient evidence to conclude that the  highlighters wrote for less than 14 continuous hours.

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