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11Alexandr11 [23.1K]
3 years ago
6

What is the probability of landing on an even number and then on a yellow section?

Mathematics
1 answer:
Ostrovityanka [42]3 years ago
6 0
<em>The answer you are looking for is: </em>
<em><u>D.) 1/4 </u></em>
<em>Hope that helps!! </em>
<em>Have a wonderful day!! </em>
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A wireless company offers two cell phone plans. For the month of September, Plan A charges $35 plus $0.25 per minute for calls.
valentina_108 [34]

Answer:

60 minutes

Step-by-step explanation:

Let the number of minutes be represented as x

For Plan A

Plan A charges $35 plus $0.25 per minute for calls.

$35 + $0.25 × x

35 + 0.25x

For Plan B

Plan B charges $20 plus $0.50 per minute for calls.

$20 + $0.50 × x

20 + 0.50x

For what number of minutes do both plans cost the same amount?

This is calculated by equating Plan A to Plan B

Plan A = Plan B

35 + 0.25x = 20 + 0.50x

Collect like terms

35 - 20 = 0.50x - 0.25x

15 = 0.25x

x = 15/0.25

x = 60 minutes.

Hence, the number of minutes that both plans cost the same amount is 60 minutes

7 0
3 years ago
The scale of the map is messing the actual distance from library to the west quill is 72 miles and it is 6 inches on the map wha
Sonbull [250]

Answer:

A) 12 miles per inch

B) 36 miles from west quail, 48 from liberty

5 0
3 years ago
A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pCI/L of radon. The resulting reading
valkas [14]

Answer:

Null hypothesis:\mu = 100  

Alternative hypothesis:\mu \neq 100  

t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921  

p_v =2*P(t_{11}  

Step-by-step explanation:

1) Data given and notation  

Data: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4

We can calculate the sample mean and deviation for this data with the following formulas:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}

The results obtained are:

\bar X=98.375 represent the sample mean  

s=0.6.109 represent the sample standard deviation  

n=12 sample size  

\mu_o =100 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 100pCL/L, the system of hypothesis are :  

Null hypothesis:\mu = 100  

Alternative hypothesis:\mu \neq 100  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

3) Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921  

4) P-value  

First we need to find the degrees of freedom for the statistic given by:

df=n-1=12-1=11

Since is a two sided test the p value would given by:  

p_v =2*P(t_{11}  

5) Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significant different from 100 at 5% of significance.  

3 0
3 years ago
Horizontal asymptote
vodomira [7]

Answer:

H.A.=4

Step-by-step explanation:

If the bottom equation is factored, you get (x+1)(x-4)

If the V.A. is -1 ( gotten by putting [x+1] equal to zero I'm guessing) then the H.A. can be gotten by putting (x-4) equal to zero.

Also, I graphed it.

6 0
3 years ago
Distribution with a mean of 100 and standard deviation of 15...
anzhelika [568]
Use the z-score:
                                      112 - 100
For 112, the z-score is --------------- = 12/15, or 0.8.  
                                             15

Next, find the AREA under the standard normal curve TO THE LEFT OF 0.8.

Use either a z-score table or a calculator for this purpose.

Using my calculator, I found that this area is 0.532.  That places 112 in the 53rd percentile.

82 would be in a lower percentile:  the 12th percentile.

Let me know if you have questions about how to use a calculator to do these calculations.

8 0
4 years ago
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