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zubka84 [21]
3 years ago
7

Allison had 310 rocks in the rock collection after three months, her collection increases by 40% how many rocks does she have no

w?
Mathematics
1 answer:
stepladder [879]3 years ago
5 0

Answer:

434

Step-by-step explanation:

310 times .4 (percentage) = 124 which is how much she gained, then you add that onto the original amount to find how many she has now, which is 434

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No, he needs to do 5(to the power of 2) plus 3(to the power of 2)
this should give him 25 plus 9 = 34
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What is the equation of its axis of symmetry?
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A Pew Internet poll asked cell phone owners about how they used their cell phones. One question asked whether or not during the
EastWind [94]

Answer:

a) \hat p=\frac{471}{1024}=0.460

The standard error is given by:

SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156

And the margin of error is given by:

ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0305

b) The 99% confidence interval would be given by (0.429;0.491)

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Data given and notation  

n=1024 represent the random sample taken    

X=471 represent the people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

\hat p=\frac{471}{1024}=0.460 estimated proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering    

p= population proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

Part a

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

The standard error is given by:

SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156

And the margin of error is given by:

ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0305

Part b

If we replace the values obtained we got:

0.460-1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.429

0.460+1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.491

The 99% confidence interval would be given by (0.429;0.491)

8 0
3 years ago
Help please with algebra problem
Strike441 [17]
Number of weekend minutes used: x
Number of weekday minutes used: y

This month Nick was billed for 643 minutes:
(1) x+y=643

The charge for these minutes was $35.44
Telephone company charges $0.04 per minute for weekend calls (x)
and $0.08 per minute for calls made on weekdays (y)
(2) 0.04x+0.08y=35.44

We have a system of 2 equations and 2 unkowns:
(1) x+y=643
(2) 0.04x+0.08y=35.44

Using the method of substitution
Isolating x from the first equation:
(1) x+y-y=643-y
(3) x=643-y

Replacing x by 643-y in the second equation
(2) 0.04x+0.08y=35.44
0.04(643-y)+0.08y=35.44
25.72-0.04y+0.08y=35.44
0.04y+25.72=35.44

Solving for y:
0.04y+25.72-25.72=35.44-25.72
0.04y=9.72

Dividing both sides of the equation by 0.04:
0.04y/0.04=9.72/0.04
y=243

Replacing y by 243 in the equation (3)
(3) x=643-y
x=643-243
x=400


Answers:
The number of weekends minutes used was 400
The number of weekdays minutes used was 243
6 0
3 years ago
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