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Kazeer [188]
3 years ago
8

Billy has 0.56 of a dollar in his pocket. Jesse has six-tenths of a dollar in his pocket. Who has the greater amount of money in

his pocket? Explain you're answer
Mathematics
2 answers:
Mumz [18]3 years ago
8 0
Jesse does. one tenth of a dollar is 10 cents, so 6 tenthas is 60 cents, which is 4 cents more than Billy has.
lana [24]3 years ago
7 0
Well first off, how good are you with decimals?
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Jina has scored 76, 85, and 86 on her previous three tests. What score does she need on her next test so that her average (mean)
Scilla [17]
Hello There!

Follow through these steps:
77 x 4 = 308
76 + 85 + 86 = 247
308 - 247 = 61.

She needs to score 61.

Hope This Helps You!
Good Luck :) 

- Hannah ❤
3 0
3 years ago
Which trigonometric function best describes the graph below?
yanalaym [24]
The function is the Sine Curve

y = sine (x)
8 0
3 years ago
Please help with 2 Math questions. I dont really understand
lisabon 2012 [21]
For the first problem, the answer is D, because every year, the graph goes down by about $4,500.
For problem two, 
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3 0
3 years ago
Which glide reflection describes the mapping ABC DEF
77julia77 [94]

Answer:

Can you provide a photo of the reflection please

Step-by-step explanation:

7 0
3 years ago
National statistics show that 23% of men smoke and 18.5% of women smoke. A random sample of 175 men indicated that 45 were smoke
NeTakaya

Answer:

98% C.I. = [-0.1270, 0.0994]

Step-by-step explanation:

Here

p1= proportion of men smokers

p1= 45/175= 0.2571

p2= proportion of women smokers

p2= 42/155= 0.2709

1)  Choose the significance level ∝= 0.02

2) The formula for confidence interval is

(p1^- p2^)- z ∝/2sqrt( p1^(1-p1^)/n1 + ( p2^(1-p2^)/n2))

and

(p1^- p2^) z ∝/2sqrt( p1^(1-p1^)/n1 + ( p2^(1-p2^)/n2))

<u>3)  The z value is 2.05</u>

<u>4) Calculations:</u>

(p1^- p2^)- z ∝/2sqrt( p1^(1-p1^)/n1 + ( p2^(1-p2^)/n2))

Putting the values

= 0.2571-0.2709)(-2.05) sqrt ( 0.2571(1-0.2571)/175+ 0.2709(1-0.2709)/155)

=-0.1270 ( using calculator)

And

(p1^- p2^)z ∝/2sqrt( p1^(1-p1^)/n1 + ( p2^(1-p2^)/n2))

Putting the values

= 0.2571-0.2709)(2.05) sqrt ( 0.2571(1-0.2571)/175+ 0.2709(1-0.2709)/155)

=0.0994  ( using calculator)

and Putting ± 2.05 in the confidence interval formula we get limits (-0.1270, 0.0994)

6 0
3 years ago
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