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BigorU [14]
3 years ago
15

Kedar is comparing the costs of phone plans. For phone plan A, the cost is $15.00 to connect and then $0.02 per minute. For phon

e plan B, the cost is $4.69 to connect and then $0.08 per minute. For what usage does plan A cost the same as plan B
Mathematics
2 answers:
sp2606 [1]3 years ago
6 0
The two plans don't fall directly on a same prive, but after 3 hours of Data usage plan B costs more than plan A

After three hours plan A= $18.60

While plan B after three hours =$19.09

That means 3 hours has to pass before plan B costs more, That is 180 minutes

The rule per hour is $1.2 per hour for plan A and $4.8 per hour for plan B 
Leto [7]3 years ago
5 0

Answer:  After 172 minutes (approx) the usage of plan A is same as the plan B.

Step-by-step explanation:

Here, For phone plan A, the cost is $15.00 to connect and then $0.02 per minute.

And, For phone plan B, the cost is $4.69 to connect and then $0.08 per minute.

Let after x minutes the total usage of plan A is same as plan B.

Therefore, 15.00 + 0.02 x = 4.69 + 0.08 x

15.00 = 4.69 + 0.08 x - 0.02 x

15.00 - 4.69 = 0.06 x

0.06 x = 10.31

x = \frac{10.31}{0.06}

x = 171.833333333\approx 172

Thus, after 172 minutes (approx) the usage of both plan will be same.

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A gondola (cable car) at a ski area holds 50 people. Its maximum safe load is 10000 pounds. A population of skiers has a distrib
fenix001 [56]

Answer:   0.03855

Step-by-step explanation:

Given :A population of skiers has a distribution of weights with mean 190 pounds and standard deviation 40 pounds.

Its maximum safe load is 10000 pounds.

Let X denotes the weight of 50 people.

As per given ,

Population mean weight of 50 people = \mu=50\times190=9500\text{ pounds}

Standard deviation of 50 people =\sigma=40\sqrt{50}=40(7.07106781187)=282.84

Then , the probability its maximum safe load will be exceeded =

P(X>10000)=P(\dfrac{X-\mu}{\sigma}>\dfrac{10000-9500}{282.84})\\\\=P(z>1.7671-8)\\\\=1-P(z\leq1.7678)\ \ \ \ [\because\ P(Z>z)=P(Z\leq z)]\\\\=1-0.96145\ \ \ [\text{ By p-value of table}]\\\\=0.03855

Thus , the probability its maximum safe load will be exceeded = 0.03855

3 0
3 years ago
Several years​ ago, 39​% of parents who had children in grades​ K-12 were satisfied with the quality of education the students r
BaLLatris [955]
<h2>Answer with explanation:</h2>

Let p be the population proportion  of parents who had children in grades​ K-12 were satisfied with the quality of education the students receive.

Given : Several years​ ago, 39​% of parents who had children in grades​ K-12 were satisfied with the quality of education the students receive.

Set hypothesis to test :

H_0: p=0.39\\\\H_a :p\neq0.39

Sample size : n= 1055

Sample proportion : \hat{p}=\dfrac{466}{1055}=0.441706161137\approx0.44

Critical value for 95% confidence : z_{\alpha/2}=1.96

Confidence interval : \hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}

0.44\pm (1.96)\sqrt{\dfrac{0.39(1-0.39)}{1055}}\\\\0.44\pm0.0299536805135\\\\0.44\pm0.03\\\\=(0.44-0.03, 0.44+0.03)\\\\=(0.41,\ 0.47)

Since , Confidence interval does not contain 0.39.

It means we reject the null hypothesis.

We conclude that 95​% confidence interval represents evidence that​ parents' attitudes toward the quality of education have changed.

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3 years ago
If ABCD is dilated by a factor of 2, the
Anna [14]
(4,4) are the coordinates of C prime
6 0
3 years ago
HELP WORTH 50 POINTS ILL GIVE U BRAINLESS IF UR RIGHT
eduard

Answer:

3.3*10^{-3}

Just like last time, you can convert them to their decimal forms, subtract, and convert back!

Hope this helps!

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4 0
3 years ago
Read 2 more answers
Is 5mi bigger or 26,300 ft
Ilya [14]
5 mi = 26400 ft.

5 mi is bigger by 100 ft.

~Evie
3 0
3 years ago
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