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oee [108]
3 years ago
13

Y'ALL! WHAT IS:46(74 + 46) + 57(9 + 36) ​

Mathematics
1 answer:
koban [17]3 years ago
6 0

Answer:

331

Step-by-step explanation:

Add 74 + 46 + 57 = 177

Add 9 + 36 = 45

Add 45 + 46 = 91

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What is 69 devided by 69<br> Can yah help and add me
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Answer:

the answer is 1

Step-by-step explanation:

69 goes into 69, 1 time

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find the area of the figure below, composed of a rectangle and semicircle. Round it to the nearest tenth.
Alexandra [31]

Answer:

Step-by-step explanation:

radius of semicircle = 5

area of semicircle = πr² = 25π

area of rectangle = 10×14 = 140

total area = 140 + 25π ≅ 218.54 square units

5 0
3 years ago
PART A: At the bakery, the daily ratio of cookie sales to donut sales is 4 to 3. Cookies and donuts cost the same amount. If the
BabaBlast [244]
Their ratio would be: Cookie / Donut = 4/3
Let, the amount of money earned from Donut = x
It would be: 4/3 = 375 / x
x = 375/4 * 3
x = 1125/4
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In short, Your Answer would be $281.25

Hope this helps!
4 0
3 years ago
Quick algebra 1 question for 10 points!
Zigmanuir [339]

Last one

Because if we simplify

  • (4x³y-6xy³)/2xy
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5 0
2 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
4 years ago
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