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mario62 [17]
2 years ago
8

Hi! I have a geometry question below, could someone please take a look?

Mathematics
1 answer:
AnnyKZ [126]2 years ago
7 0

Answer:

109 degrees.

Step-by-step explanation:

m< AEB = 1/2 (98 + 120)

= 1/2 * 218

= 109 degrees.

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( NEEDED ASAP! )
bulgar [2K]

Answer:

y=3r+5

Step-by-step explanation:

Break up the sentance.

Capon city has 5 more than 3 times as many residents as harbor city.

The sentance above can be written like this: y=5+3x. Since they want r to represent the number of people in Harbor city, use r in place of x.

Capon city=y

has=equal sign

5 more=+5

3 times as many=3r

y=3r+5

4 0
2 years ago
PLEASE HELP ME PLEASE!!
Harman [31]
  1. Answer:

hypotenuse (h)=?

perpendicular (p)=3

base(b)=2

using pythagoras theorem

h²=p²+b²

h²=3²+2²

h²=9+4

h²=13

h=√13

  • hypotenuse (h)=12
  • perpendicular (p)=10
  • base(b)=?

using pythagoras theorem

b²=h²-p²

b²=12²-10²

b²=144-100

b=√44

b=2√11

6 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
The SAT mathematics scores in the state of Florida are approximately normally distributed with a mean of 500 and a standard devi
Yanka [14]
The interval (300,700) corresponds to the part of the distribution lying within 2 standard deviations of the mean (since 500-2*100=300 and 500+2*100=700). The empirical rule states that approximately 95% of the distribution is expected to fall in this range.
7 0
3 years ago
Read 2 more answers
Can someone help me w/ these please? Theyre seperate questions so there will be seperate answers... and i need to show work
Sergeeva-Olga [200]
The first figure below shows the graph for problem 5).

The second figure below shows the graph for problem 6).

7 0
2 years ago
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