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almond37 [142]
4 years ago
6

HELP QUICK IM GIVING 21 POINTS AND A BRAINLIEST AWARD

Mathematics
2 answers:
Andrei [34K]4 years ago
6 0

Answer:

10:45

Step-by-step explanation:

v...speed = 40 mph

s...distance = 60

t...time

if he was 15 min late he arrived at 12:15

v=s/t => t=s/v

t = 60/40 = 1.5 h = 1h and 30 min

if he arrived at 12:15 and he was driving for an hour and a half just substract

12:15 - 1:30 = 11:15 - 0:30 = 10:45

he left at 10 45

iren [92.7K]4 years ago
5 0
V...speed = 40 mph
s...distance = 60
t...time

if he was 15 min late he arrived at 12:15

v=s/t => t=s/v
t = 60/40 = 1.5 h = 1h and 30 min

if he arrived at 12:15 and he was driving for an hour and a half just substract
12:15 - 1:30 = 11:15 - 0:30 = 10:45

he left at 10 45



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16 years

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1.7% of 2000 is $34

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Which equation has the same graph as 4x + 8y = 12?
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Calculate volume of hemisphere with radius 3.2 cm
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Answer:

V ≈ 68.63 cm³

Step-by-step explanation:

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Nate has $50 to spend at the grocery store. he fills his shopping cart with items totaling $46. At checkout he will have to pay
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Read 2 more answers
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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