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sveta [45]
3 years ago
5

Alexis G. has a cell phone plan that charges $0.05 per minute plus a monthly fee of $25.00. She budgets $35.50 per month for tot

al cell phone expenses without taxes. What is the maximum number of minutes Alexis could use her phone each month in order to stay within her budget?
NEED ANSWER IMIDIETLY!!!!!!!!!!!!
Mathematics
1 answer:
jeyben [28]3 years ago
6 0
The Maximum amount of minutes she could use would be 210

Formula:
0.05x + 25 <= 35.50
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The radius of a circle is 4.53 inches what is the length of the circumference of the circle ?
gtnhenbr [62]

Answer:

28.46in

Step-by-step explanation:

C=2πr

C=2π*4.53

C≈28.46in

8 0
3 years ago
Sonji paid $25 for two scarves, which were different prices. When she got home she could not find the receipt. She remembered th
Oksanka [162]

Answer:

$14

Step-by-step explanation:

Make the variables:

x

x + 3

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x = 11

11, 14

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3 years ago
Read 2 more answers
What is 8833/10000 in simplest form
tresset_1 [31]
Well if you're asking for it to be in a fraction, then it's already in its simpliest form. but for a decimal, it would be 0.8833
4 0
3 years ago
An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de
Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}

Then we have:

P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}

Then we have

P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453

The sum of the probabilities is

P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981

8 0
3 years ago
I don’t know what to do
LuckyWell [14K]

original points (-2,0), (2,0), (0,2)

new points (-6.25,0),(6.25,0),(0,6.25)

the point (0,6.25) is the vertex that lies on y -axis

8 0
3 years ago
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