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frozen [14]
3 years ago
11

Last week Rachel and her roommates ate 1/3 of a carton of yogurt, and this week they ate 2/3 of a carton. How much more yogurt d

id they eat this week compared to last week?
Mathematics
2 answers:
stellarik [79]3 years ago
8 0
1/3 more yogurt than last week
liraira [26]3 years ago
5 0

Answer: 1/3 more

Step-by-step explanation:

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The answer would be 84
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A rectangle is drawn on a coordinate grid. three of its four vertices are located points ( -1,-2), (-1,4) and (2,-2). What is th
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C=(2,4) is the vertex missing.
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Identify the asymptotes of y=1 over x -6<br><br><br> A.<br> B.<br> C.<br> D.
Westkost [7]

Answer:

  D.  x = 0; y = -6

Step-by-step explanation:

There is a vertical asymptote at the value of x that makes the denominator zero. That value is x=0. This is sufficient to choose the correct answer.

When x gets large, the term 1/x nears zero, so the value of y nears -6. This is the horizontal asymptote.

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A bag contains three red marbles, four blue, nine yellow, and five green. what is the probability of getting a yellow or green
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Step-by-step explanation:

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A study of 4,430 U.S. adults found that 2,913 were obese or overweight. The 95% confidence interval for the proportion of adults
IrinaK [193]

Answer:

0.658 - 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.644

0.658 + 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.672

We are 95% confident that the true proportion of adults overweight is between 0.644 and 0.672

Step-by-step explanation:

We can begin founding the estimated proportion of people overweight:

\hat p =\frac{2913}{4430}= 0.658

We need to find a critical value for the confidence level using the normal standard distribution. We know that 95% is the confidence level, then the significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The confidence interval for the true population proportion of interest is given by the following formula:  

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

Replacing the data provided we got:

0.658 - 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.644

0.658 + 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.672

We are 95% confident that the true proportion of adults overweight is between 0.644 and 0.672

6 0
3 years ago
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