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kakasveta [241]
3 years ago
11

Dakota makes a salad dressing by Combining 7 1/3 fluid ounces of oil and 2 1/8 fluid ounces of vinegar in a jar. she then pours

2 1/4 fluid ounces of the dressing on to her salad. How much dressing remains in the jar?
Mathematics
1 answer:
pshichka [43]3 years ago
6 0
7 1/3 + 2 1/8 = 9 11/24

9 11/24 - 2 1/4= 7 5/24

Your answer is 7 5/24


Hope I helped!

Let me know if you need anything else!

<span>~ Zoe</span>
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Solve -52 = 4m for m
ZanzabumX [31]

Note the equal sign. What you do to one side, you do to the other.


Isolate the variable. Divide 4 from both sides


(-52)/4 = (4m)/4


m = -52/4


m = -13


-13 is your answer for m



hope this helps

8 0
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It's easy I'm just not I'm my head rn can someone help?
Dominik [7]
Answer: all of them
5 0
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I don’t know what the last one is
joja [24]

But its still not that much info what is the answers for it if there is no answer

How are you finding what does x equal

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Which symbol is used to name a line with two points A and B
Mars2501 [29]

Answer:

AB

Step-by-step explanation:

I think that's the answer

5 0
3 years ago
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The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college f
Sloan [31]

Answer:

Step-by-step explanation:

Hello!

There are two variables of interest:

X₁: number of college freshmen that carry a credit card balance.

n₁= 1000

p'₁= 0.37

X₂: number of college seniors that carry a credit card balance.

n₂= 1000

p'₂= 0.48

a. You need to construct a 90% CI for the proportion of freshmen  who carry a credit card balance.

The formula for the interval is:

p'₁±Z_{1-\alpha /2}*\sqrt{\frac{p'_1(1-p'_1)}{n_1} }

Z_{1-\alpha /2}= Z_{0.95}= 1.648

0.37±1.648*\sqrt{\frac{0.37*0.63}{1000} }

0.37±1.648*0.015

[0.35;0.39]

With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.

b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648

The formula for this interval is

p'₂±Z_{1-\alpha /2}*\sqrt{\frac{p'_2(1-p'_2)}{n_2} }

0.48±1.648* \sqrt{\frac{0.48*0.52}{1000} }

0.48±1.648*0.016

[0.45;0.51]

With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.

c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.

Freshmen: \sqrt{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015

Seniors: \sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{0.48*0.52}{1000} }= 0.01579 = 0.016

The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.

8 0
3 years ago
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