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vlada-n [284]
3 years ago
15

A bag of dog food says that it now contains 20% more. Originally, it came with 63 ounces of dog food. How much dog food does the

bag come with now? Set up the double number line to show that 63 ounces is 100%.
Mathematics
1 answer:
Brrunno [24]3 years ago
5 0

Answer:

75.6 ounces

Step-by-step explanation:

Step one.

Given data

we are told that the original quantity of food is 63 ounces

and the increase is in food is by 20%

Step two:

let us find the increase in ounces

=20/100*63

=0.2*63

=12.6 ounces

Hence the amount of food in the bag now is

=12.6+63

=75.6 ounces

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-6x + 7 = 8x + A

let x = -4
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24 + 7      = -32 + A
31 + 32    = -32 + 32 + A
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A = 63
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What is a the gcf of 8
liraira [26]

Answer:

Step-by-step explanation:

4

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Mr. Burrows has cups of flour.
Alexeev081 [22]

The question is incomplete :

However, assume Number of cups of flower 10

Number of flour cups for each pie 1/4

Answer:

8

Step-by-step explanation:

Number of cups of flower = 10

Number of flower cups per crust pie = 1/4 cups

Since, the cups of flower was divided equally into to 5 containers;

Number of flower cups per container :

10 / 5 = 2

Each container has 2 cups of flower each.

1 container was used in making crust pie :

The number of crust pies made is:

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8 0
3 years ago
Suppose that administrators at a large urban high school want to gain a better understanding of the prevalence of bullying withi
boyakko [2]

Answer:

lower limit =0.261

upper limit =0.369

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Description in words of the parameter p

p represent the real population proportion of people that have experienced bullying

X= 63 people in the random sample that have experienced bullying

n=200 is the sample size required  

\hat p=\frac{63}{200}=0.315 represent the estimated proportion of people that have experienced bullying

z_{\alpha/2} represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Numerical estimate for p

In order to estimate a proportion we use this formula:

\hat p =\frac{X}{n} where X represent the number of people with a characteristic and n the total sample size selected.

\hat p=\frac{63}{200}=0.315 represent the estimated proportion of people that they were planning to pursue a graduate degree

Confidence interval

The confidence interval for a proportion is given by this formula  

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

For the 90% confidence interval the value of \alpha=1-0.90=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.64  

And replacing into the confidence interval formula we got:  

0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.261  

0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.369  

And the 90% confidence interval would be given (0.261;0.369).  

We are confident at 90% that the true proportion of people that they were planning to pursue a graduate degree is between (0.261;0.369).  

lower limit =0.261

upper limit =0.369

5 0
4 years ago
Enter the correct answer in the box.
Alex

Answer:

-2x+1 ez clappppppppppppppppppppp

7 0
2 years ago
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