Number 8 fill in the blanks

Four friends go to the movies and pay $11 per ticket. They also each buy a snack

Setler79 [48]

Answer:

Each snack combo costed $8

Step-by-step explanation:

4 friends* 11 per ticket = $44 for four tickets

76 - 44 = $32 spent on all the snack combos

$32 ÷ 4 friends = $8 spent on each snack combo

So, each snack combo costed $8

Hope this helps!

6 0

2 years ago

Read 2 more answers

A builder could get 9 sheets of Sheetrock for $20. If he bought 15 sheets, how much money would he have spent?

lora16 [44]

$33.33

20/9 = $2.22 per sheet
$2.22*15 = $33.33

4 0

3 years ago

Read 2 more answers

The rectangular bar is connected to the support bracket with a 10-mm-diameter pin. The bar width is w = 70 mm and the bar thickn

german

Answer:

P_max = 9.032 KN

Step-by-step explanation:

Given:

- Bar width and each side of bracket w = 70 mm

- Bar thickness and each side of bracket t = 20 mm

- Pin diameter d = 10 mm

- Average allowable bearing stress of (Bar and Bracket) T = 120 MPa

- Average allowable shear stress of pin S = 115 MPa

Find:

The maximum force P that the structure can support.

Solution:

- Bearing Stress in bar:

T = P / A

P = T*A

P = (120) * (0.07*0.02)

P = 168 KN

- Shear stress in pin:

S = P / A

P = S*A

P = (115)*pi*(0.01)^2 / 4

P = 9.032 KN

- Bearing Stress in each bracket:

T = P / 2*A

P = T*A*2

P = 2*(120) * (0.07*0.02)

P = 336 KN

- The maximum force P that this structure can support:

P_max = min (168 , 9.032 , 336)

P_max = 9.032 KN

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

A True/False quiz has three questions. When guessing, the probability of getting a question correct is the same as the probabili

rewona [7]

Answer:

iycdjrasgmbjbljvkg

Step-by-step explanation:

xgcuchjlhgjfa1347722j

6 0

3 years ago