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RSB [31]
2 years ago
5

BRAINLIEST! FAST ! PLEASE BE SURE

Mathematics
2 answers:
vaieri [72.5K]2 years ago
7 0

Answer:

-7

Step-by-step explanation:

(1/2)2−6(2−2/3)

1- 6(4/3)

1- (24/3)

-7

rosijanka [135]2 years ago
7 0

Answer:

-\frac{31}{8}

Step-by-step explanation:

We are given that an expression

(\frac{1}{2})^2-6(2-\frac{2}{3})

We have to find the value as a simplified fraction

\frac{1}{2}\times \frac{1}{2}-6(\frac{6-2}{3})

\frac{1}{4}-6\times\frac{4}{3}

\frac{1}{4}-2\times 4

\frac{1}{4}-8

\frac{1-32}{8}

-\frac{31}{8}

Hence, (\frac{1}{2})^2-6(2-\frac{2}{3})=-\frac{31}{8}

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Practice using the work formula. A bathtub that holds 42 gallons of water is filling at a rate of 4 gallons per minute. How long
trapecia [35]

Answer: The percentage of the tub after 6 minutes is 30 %

Step-by-step explanation: The  calculation can be done as follows

1 minutes= 4 gallons

6 minutes= 4×6

= 24 minutes

If the bathtub can hold 80 gallons of water then the percentage of the tub after 6 minutes can be calculated as follows

= 24/80

= 0.3 × 100

= 30 %

Hence the percentage of the tub will be filled after 6 minutes is 30 %

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2 years ago
Which is less, 8 ½ pounds or 130 ounces?
Harman [31]
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PLEASE HELP 9 THROUGH<br> 12 ILL MARK BRAINLEST
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chet needs to buy 4 work shirts,all costing the same amount .the total cost before chet appplies a $25 gift certificate can be n
Rashid [163]
So for this problem, let us use x as the cost before Chet would apply a $25 gift certificate. Based on the problem, we can see that the original cost of the product cannot be more than 75 which means that it can be equal to 75 or less than 75. We can actually express the inequality as x< or = 75 since we are looking for the cost before Chet applied the $25 gift certificate. This means that we do not need to add in the 25 yet since the question asks for the cost before the application of the discount. 
5 0
3 years ago
5) In a certain supermarket, a sample of 60 customers who used a self-service checkout lane averaged 5.2 minutes of checkout tim
Ne4ueva [31]

Answer:

\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643

S_p=2.940

t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751

df=60+72-2=130

p_v =P(t_{130}

Assuming a significance level of \alpha=0.05 we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

Step-by-step explanation:

Data given

Our notation on this case :

n_1 =60 represent the sample size for people who used a self service

n_2 =72 represent the sample size for people who used a cashier

\bar X_1 =5.2 represent the sample mean for people who used a self service

\bar X_2 =6.1 represent the sample mean people who used a cashier

s_1=3.1 represent the sample standard deviation for people who used a self service

s_2=2.8 represent the sample standard deviation for people who used a cashier

Assumptions

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

And t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

System of hypothesis

Null hypothesis: \mu_1 \geq \mu_2

Alternative hypothesis: \mu_1 < \mu_2

This system is equivalent to:

Null hypothesis: \mu_1 - \mu_2 \geq 0

Alternative hypothesis: \mu_1 -\mu_2 < 0

We can find the pooled variance:

\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643

And the deviation would be just the square root of the variance:

S_p=2.940

The statistic is given by:

t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751

The degrees of freedom are given by:

df=60+72-2=130

And now we can calculate the p value with:

p_v =P(t_{130}

Assuming a significance level of \alpha=0.05 we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

5 0
3 years ago
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