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Andreas93 [3]
3 years ago
8

dante’s cell phone company charges $45 per month for unlimited calls and internet use and $0.25 per text message. his last cell

phone bill was $60.50. how many text messages did he send last month. write an equation.
Mathematics
2 answers:
anastassius [24]3 years ago
6 0

Answer:

62 text messages

Step-by-step explanation:

Let the number of text messages be x,

<u>Then the required equation:</u>

  • 0.25x + 45 = 60.50

<u>Solution:</u>

  • 0.25x = 60.50 - 45
  • 0.25x = 15.50
  • x = 15.50/0.25
  • x = 62

<u>Answer is</u>: 62 text messages

dybincka [34]3 years ago
5 0

Answer:

242 text messages.

Step-by-step explanation:

60.50 / 0.25 = 242

60.50 being the phone bill.

0.25 being each text message.

/ stands for ' divided by '.

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Solution :

The test is distributed normally with mean of 72.8 and the standard deviation of 7.3

Finding numerical limits for the D grade.

D grade : Scores below the top 80% and above the bottom 10%.

Let the bottom limit for D grade be $D_1$ and the top limit for D grade be $D_2$.

First find the bottom numerical limit for a D grade is :

$P(X

$P(X\leq D_1)= 0.10$

$P\left(\frac{X-\mu}{\sigma} \leq \frac{D_1-\mu}{\sigma}\right) = 0.10$

$P\left(Z \leq \frac{D_1-72.8}{7.3}\right) = 0.10$    ..........(1)

From (1)

$\frac{D_1 - 72.8}{7.3} = -1.28$

$D_1 = -1.28(7.3)+72.8$

      = 63.45

       ≈ 64

Now the top numerical limit for D grade :

$P(X>D_2)= 0.80$

$1-P(X\leq D_2)= 0.80$

$P(X\leq D_2)= 1-0.80$

$P(X\leq D_2)= 0.20$

$P\left(\frac{X-\mu}{\sigma} \leq \frac{D_2-\mu}{\sigma}\right) = 0.20$

$P\left(Z \leq \frac{D_2-72.8}{7.3}\right) = 0.20$    ..........(2)

From (2)

$\frac{D_2- 72.8}{7.3} = -0.84$

$D_12= -0.84(7.3)+72.8$

      = 66.668

       ≈ 67

Therefore, the numerical limit for a D grade is 64 to 67.

7 0
3 years ago
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