What Is the property of (6 +56) + 5 = (56 + 6) + 5

Mabel paid $2.95 for a gronola bar with dimes and quarters. she used 5 fewer quaters then dimes. how many dimes and quarters did

Oduvanchick [21]

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7 0

3 years ago

4(8x+6)-2=22 Please help

VMariaS [17]

Answer:

x=0

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Please help will give brainliest

SVEN [57.7K]

Answer:

The diameter is: 8 feet

The radius is: 4 ft.

The area is: 50.27

Step-by-step explanation:

Hope this helps!

8 0

3 years ago

Solve each system using substitution: 1. 4x - 5y= -15 y= -3x + 22

Bumek [7]

4x - 5y = -15
y = -3x + 22

Plug in your y= equation into the y variable in the other equation

4x - 5 (y) = -15
^^
y = (-3x + 22) = -15

4x - 5 (-3x + 22) = -15

Distribute

4x + 15x - 110 = -15

Combine like terms and add 110 over to -15

19x = 95

Then, divide the whole equation by 19

x = 5

Then, plug in your x into your y= equation

y = -3 (5) + 22
y = -15 + 22
y = 7

x = 5
y = 7
(5, 7)

3 0

3 years ago

A person purchased a slot machine and tested it by playing it 1,137 times. There are 10 different categories of outcomes, includ

ivann1987 [24]

Answer:

$p_v= P(\chi^2_{9}>11.517)=0.2419$

And on this case if we see the significance level given $\alpha=0.1$ we see that $p_v>alpha$ so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.

Step-by-step explanation:

A chi-square goodness of fit test determines if a sample data obtained fit to a specified population.

$p_v$ represent the p value for the test

O= obserbed values

E= expected values

The system of hypothesis for this case are:

Null hypothesis: $O_i = E_i[/tex[Alternative hypothesis: [tex]O_i \neq E_i$

The statistic to check the hypothesis is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

On this case after calculate the statistic they got: $\chi^2 = 11.517$

And in order to calculate the p value we need to find first the degrees of freedom given by:

$df=n-1=10-1=9$ , where k represent the number of levels (on this cas we have 10 categories)

And in order to calculate the p value we need to calculate the following probability:

$p_v= P(\chi^2_{9}>11.517)=0.2419$

And on this case if we see the significance level given $\alpha=0.1$ we see that $p_v>alpha$ so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.

6 0

3 years ago