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3 years ago

13

ok the 100 point question is... if i am walking to work at 6:00 and it takes me three hours to get there, and jimmy walks to the

same place of work two hours later at 8:00 walking two mph faster than me, what time did i get to work?

2 answers:

wariber [46]3 years ago

4 0

9:00 tricky question also

makkiz [27]3 years ago

3 0

Ok so at 6:00, you start walking. It takes me 3 hours to get there
6+3=9
you got there at 9:00
no idea what the jimmy buisieness is

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El área de un rectángulo es 6 x al cuadrado más 2x en cm Determine su base y altura

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Answer:

Tres respuestas posibles son:

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y hay muchas mas.

Step-by-step explanation:

Hay muchas respuestas posibles.

6x^2 + 2x = 2(3x^2 + x) = 2x(3x + 2) = x(6x + 2)

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Each sister in the Johnson family owns one of these fish tanks. Each tank is shaped like a rectangular prism. They want to combi

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Answer:

B.)536 $cm^{3}$

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Find the volume of all the fish tanks.

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3 years ago

A manager of a grocery store wants to determine if consumers are spending more than the national average. The national average i

strojnjashka [21]

The valid conclusions for the manager based on the considered test is given by: Option

<h3>When do we perform one sample z-test?</h3>

One sample z-test is performed if the sample size is large enough (n > 30) and we want to know if the sample comes from the specific population.

For this case, we're specified that:

Population mean = $\mu$ = $150
Population standard deviation = $\sigma$ = $30.20
Sample mean = $\overline{x}$ = $160
Sample size = n = 40 > 30
Level of significance = $\alpha$ = 2.5% = 0.025
We want to determine if the average customer spends more in his store than the national average.

Forming hypotheses:

Null Hypothesis: Nullifies what we're trying to determine. Assumes that the average customer doesn't spend more in the store than the national average. Symbolically, we get: $H_0: \mu_0 \leq \mu = 150$
Alternate hypothesis: Assumes that customer spends more in his store than the national average. Symbolically $H_1: \mu_0 > \mu = 150$

where $\mu_0$ is the hypothesized population mean of the money his customer spends in his store.

The z-test statistic we get is:

$z = \dfrac{\overline{x} - \mu_0}{\sigma/\sqrt{n}} = \dfrac{160 - 150}{30.20/\sqrt{40}} \approx 2.094$

The test is single tailed, (right tailed).

The critical value of z at level of significance 0.025 is 1.96

Since we've got 2.904 > 1.96, so we reject the null hypothesis.

(as for right tailed test, we reject null hypothesis if the test statistic is > critical value).

Thus, we accept the alternate hypothesis that customer spends more in his store than the national average.

Learn more about one-sample z-test here:

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Solving a Two-Step Matrix Equation<br> Solve the equation:

Cloud [144]

Answer:

$\boxed {x_{1} = 3}$

$\boxed {x_{2} = -4}$

Step-by-step explanation:

Solve the following equation:

$\left[\begin{array}{ccc}3&2\\5&5\\\end{array}\right] \left[\begin{array}{ccc}x_{1}\\x_{2}\\\end{array}\right] + \left[\begin{array}{ccc}1\\2\\\end{array}\right] = \left[\begin{array}{ccc}2\\-3\\\end{array}\right]$

-In order to solve a pair of equations by using substitution, you first need to solve one of the equations for one of variables and then you would substitute the result for that variable in the other equation:

-First equation:

$3x_{1} + 2x_{2} + 1 = 2$

-Second equation:

$5x_{1} + 5x_{2} + 2 = -3$

-Choose one of the two following equations, which I choose the first one, then you solve for $x_{1}$ by isolating

$3x_{1} + 2x_{2} + 1 = 2$

-Subtract $1$ to both sides:

$3x_{1} + 2x_{2} + 1 - 1 = 2 - 1$

$3x_{1} + 2x_{2} = 1$

-Subtract $2x_{2}$ to both sides:

$3x_{1} + 2x_{2} - 2x_{2} = -2x_{2} + 1$

$3x_{1} = -2x_{2} + 1$

-Divide both sides by $3$ :

$3x_{1} = -2x_{2} + 1$

$x_{1} = \frac{1}{3} (-2x_{2} + 1)$

-Multiply $-2x_{2} + 1$ by $\frac{1}{3}$ :

$x_{1} = \frac{1}{3} (-2x_{2} + 1)$

$x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}$

-Substitute $-\frac{2x_{2} + 1}{3}$ for $x_{1}$ in the second equation, which is $5x_{1} + 5x_{2} + 2 = -3$ :

$5x_{1} + 5x_{2} + 2 = -3$

$5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3$

Multiply $-\frac{2x_{2} + 1}{3}$ by $5$ :

$5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3$

$-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3$

-Combine like terms:

$-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3$

$\frac{5}{3}x_{2} + \frac{11}{3} = -3$

-Subtract $\frac{11}{3}$ to both sides:

$\frac{5}{3}x_{2} + \frac{11}{3} - \frac{11}{3} = -3 - \frac{11}{3}$

$\frac{5}{3}x_{2} = -\frac{20}{3}$

-Multiply both sides by $\frac{5}{3}$ :

$\frac{\frac{5}{3}x_{2}}{\frac{5}{3}} = \frac{-\frac{20}{3}}{\frac{5}{3}}$

$\boxed {x_{2} = -4}$

-After you have the value of $x_2$ , substitute for $x_{2}$ onto this equation, which is $x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}$ :

$x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}$

$x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}$

-Multiply $-\frac{2}{3}$ and $-4$ :

$x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}$

$x_{1} = \frac{8 + 1}{3}$

-Since both $\frac{1}{3}$ and $\frac{8}{3}$ have the same denominator, then add the numerators together. Also, after you have added both numerators together, reduce the fraction to the lowest term:

$x_{1} = \frac{8 + 1}{3}$

$x_{1} = \frac{9}{3}$

$\boxed {x_{1} = 3}$

5 0

3 years ago