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2 years ago

14

Lauren wants to keep her cell phone bill under $60 per month. Her current cell phone plan is $30 per month plus $0.05 per text.

Write and inequality to represent the number of texts, , Lauren can send each month while staying within her budget.

1 answer:

KiRa [710]2 years ago

7 0

Answer:

Step-by-step explanation:

Let the number of text messages be x

30 + 0.05x <= 60

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Which fraction is equivalent to the fraction shown by this model. 12/9

svp [43]

Answer:

4/3

Step-by-step explanation:

12/9 is equivalent to 4/3, because you can divide both the numerator and denominator by the common factor of 3 and that will give you the correct fraction: 4/3.

Hope this helped.

5 0

2 years ago

A pet store contains 35 light green parakeets (14 females and 21 males) and 44 sky blue parakeets (28 females and 16 males). You

snow_tiger [21]

You can use this formula <span>P(AorB) = P(A) + P(B) - P(AandB)

Given:
35 LG (14 F & 21 M)
44 SB (28 F & 16 M)

Req:
- the probability that it is a female (F) or a sky blue (SB)

Sol:
</span>P(F or SB) = P(F) + P(SB) - P(F and SB)
= [(14 F + 28 F)/(35 + 44)] + [(44 SB)/(35 + 44)] - [(28 F)/(35 + 44)]
= 53.16 + 55.70 - 35.44
= 73.42%

You have to deduct 28 female parakeets from 44 sky blue parakeets because the 28 parakeets are already accounted for in the female parakeets. You can also think of how many ways you can choose a female parakeet and a sky blue parakeet. Add all female parakeets (14 + 28) = 42. Sky blue parakeet equaled to 44. Minus the 28 female parakeets included in the sky blue parakeet to avoid double counting. 42 + 44 - 28 = 58 divided by 79 (35 + 44) total parakeets = 73.42%

7 0

2 years ago

A deck of 52 cards contains 12 picture cards. If the 52 cards are distributed in a random manner among four players in such a wa

Mkey [24]

Answer:

The probability that each player will receive three picture cards = 0.0324

Step-by-step explanation:

As given,

A deck of 52 cards contains 12 picture cards

Remaining card = 52 - 12 = 40

So,

Total number of ways in which 12 picture card is distributed = $\frac{12!}{3! 3! 3! 3!}$

Now,

The Total number of ways in which Remaining cards are distributed = $\frac{40!}{10! 10! 10! 10!}$

So,

Total number of ways of getting 3 picture card and remaining card = $\frac{12!}{3! 3! 3! 3!}$ × $\frac{40!}{10! 10! 10! 10!}$

= $\frac{12! 40!}{(3!)^{4} (10!)^{4} }$

Now,

Total number of ways to distribute 52 cards so that each people get 13 card = $\frac{52!}{13! 13! 13! 13!} = \frac{52!}{ (13!)^{4} }$

∴ The probability = $\frac{\frac{12! 40!}{(3!)^{4} (10!)^{4} }}{\frac{52!}{(13!)^{4} }}$

= $\frac{12! 40!}{(3!)^{4} (10!)^{4} }$ × $\frac{(13!)^{4} }{ 52! }$

= $\frac{12! 40!}{(3!)^{4} (10!)^{4} }$ × $\frac{(13.12.11.10!)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41.40! }$

= $\frac{12!}{(3!)^{4} }$ × $\frac{(13.12.11)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41}$

= $\frac{479,001,600}{(6)^{4} }$ × $\frac{(1716)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41}$

= 0.0324

∴ we get

The probability that each player will receive three picture cards = 0.0324

6 0

2 years ago

PT= 5x+3 and TQ= 48 midpoint of PQ is

prisoha [69]

Remark

This really can't be done unless you know that T is the midpoint. I'm pretty sure it is intended to be.

Solve

5x + 3 = 48 Subtract 3

5x = 48 - 3

5x = 45 Divide by 5

x = 45/5

x = 9

8 0

2 years ago

Find the gradient of the straight line 2y+3x=0

MArishka [77]

3x + 2y = 0
- 3x - 3x
2y = -3x
2 2
y = -1.5x

The gradient of the straight line of the function is -1.5.

6 0

2 years ago