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3 years ago

What is the solution to this equation?

Mathematics

1 answer:

Radda [10]3 years ago

5 0

Answer:

9. This is because you have to add 13 to -4, which gives you positive 9. Hope this helped!

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Write an equation for the line below

Stells [14]

The answer is : y = -3x + 1

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3 years ago

Write the expression without the fraction bar.

kirill115 [55]

For problems like this you can move the y's on the bottem up top by fliping the sign of the exponent (in this case 7 to -7) and MULTIPLYING it with EVERYTHING on top, then because of properties of multplication you can add the exponents to combine them into one term (in this case add 3 and -7 to get -4) $\frac{ y^{3} }{y^{7}}=y^3 *y^{-7} = y^{3+-7}=y^{3-7}=y^-4$ If any of this does not make sense let me know and ill try to clarify better

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Complete the following by filling in the blanks.

max2010maxim [7]

Answer:

0.5 x 2 = 1 so therefore area height base = 1in

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3 years ago

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Solve x2 + 1/2x +1/16=4/9

kolezko [41]

<u>Answer:</u>

x = 0.417 or x = -0.917

<u>Step-by-step explanation:</u>

We are given the following expression and we are to solve it for the variable x:

$x ^2 + \frac { 1 } { 2 } x + \frac { 1 } { 1 6 } = \frac { 4 } { 9 }$

We will find the least common multiple of 2. 6 and 9:

$x^2 \times 144 +\frac{1}{2}x \times 144 +\frac{1}{16} \times 144 = \frac{4}{9} \times 144$

Simplifying it to get:

$144x^2+72x+9=64$

$144x^2+72x+9-64=0$

$144x^2+72x-55=0$

Using the quadratic formula to solve for x:

$x=\frac{-b \pm\sqrt{b^2-4ac} }{2a}$

$x=\frac{-72 \pm\sqrt{72^2-4(144)(-55)} }{2a}$

$x=\frac{5}{12}$ or $0.417$

$x=-\frac{11}{12}$ or $-0.917$

3 0

3 years ago

An educator claims that the average salary of substitute teachers in school districts is less than $60 per day. A random sample

valentina_108 [34]

Answer:

$t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626$

The degrees of freedom are given by:

$df=n-1=8-1=7$

The p value would be given by:

$p_v =P(t_{(7)}$

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

Step-by-step explanation:

Information given

60, 56, 60, 55, 70, 55, 60, and 55.

We can calculate the mean and deviation with these formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

Replacing we got:

$\bar X=58.875$ represent the mean

$s=5.083$ represent the sample standard deviation for the sample

$n=8$ sample size

$\mu_o =60$ represent the value that we want to test

$\alpha=0.1$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean is less than 60, the system of hypothesis would be:

Null hypothesis: $\mu \geq 60$

Alternative hypothesis: $\mu < 60$

The statistic would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info we got:

$t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626$

The degrees of freedom are given by:

$df=n-1=8-1=7$

The p value would be given by:

$p_v =P(t_{(7)}$

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

3 0

3 years ago