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rusak2 [61]
2 years ago
13

Which of the following are equivalent to the expression 9/4 ( 1.75 - 2 1/4)? Select all that apply.

Mathematics
1 answer:
Sindrei [870]2 years ago
4 0

Answer:

9/4 (1.75 - 21/4)

=9/4 (1.75 - 5.25)

=2.25 -3.5

= -1.25

I think your f answer is wrong

there should be -1.25

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This is the dumbest experiment in the world. Like seriously _s o u p c a n s ?

8 0
3 years ago
What are two decimals equivalent to 9.1
Marizza181 [45]
To get equivalent decimals you times the decimal by a number for example 2 
9.1x2= 18.2
or times it by 3 
9.1x3= 27.3
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i hope this helped :)
8 0
3 years ago
1250 / 16 show work please answer fast
OLga [1]

Answer:

78.125

Step-by-step explanation:

sorry i can show the work :(

5 0
2 years ago
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
Giving best answer the brainliest answer ya hurddddddddddddddd!
Genrish500 [490]
4th answer is the answer for theis question because they are equivalent
7 0
3 years ago
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