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Ahat [919]
2 years ago
5

A recipe calls for 1/4 cup of broth. The largest volume-measuring tool that Tabitha has is a teaspoon. She knows that there are

16 tablespoons in 1 cup and 3 teaspoons in 1 tablespoon. How many teaspoons of broth should she use?
a) 1 1/3 teaspoons
b) 12 teaspoons
c) 19 teaspoons
d) 21 1/3 teaspoons
Mathematics
1 answer:
Lady bird [3.3K]2 years ago
7 0

Answer:

Option b. 12 teaspoons.

Step-by-step explanation:

In the question it given :

1 cup = 16 tablespoons

1 tablespoon = 3 teaspoons

16 tablespoon = 48 teaspoons or 1 cup

A recipe calls for 1/4 cup of broth. Tabitha has only a teaspoon to measure broth.

1 cup broth = 48 teaspoons

1/4 cup broth = 48 × \frac{1}{4} = 12 teaspoons

Tabitha should use 12 teaspoons of broth.

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What is the length of the radius?

12 units

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Scott wanted to bjy a sofa which originally cost $500.Store A was selling it for 20% off plus 6.5% sales tax.Store B was selling
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Answer:

Store A offers the least amount for the sofa

Step-by-step explanation:

Cost of sofa =$500

For store A

Discount =20% off

The amount of the discount is

=20/100*500

=0.2*500

=$100

6.5% sales tax

The amount of tax is

=6.5/100*500

=0.065*500

=$32.5

Total cost of the sofa

=500-100+32.5

=400+32.5

=$432.5

For store B

Discount =30% off

The amount of the discount is

=30/100*500

=0.3*500

=$150

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Total cost of sofa

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3 0
2 years ago
Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a
kipiarov [429]

Answer:

a) P(k≤11) = 0.021

b) P(k>23) = 0.213

c) P(11≤k≤23) = 0.777

P(11<k<23) = 0.699

d) P(15<k<25)=0.687

Step-by-step explanation:

a) What is the probability that the number of drivers will be at most 11?

We have to calculate P(k≤11)

P(k\leq11)=\sum_0^{11} P(k

P(k=0) = 20^0e^{-20}/0!=1 \cdot 0.00000000206/1=0\\\\P(k=1) = 20^1e^{-20}/1!=20 \cdot 0.00000000206/1=0\\\\P(k=2) = 20^2e^{-20}/2!=400 \cdot 0.00000000206/2=0\\\\P(k=3) = 20^3e^{-20}/3!=8000 \cdot 0.00000000206/6=0\\\\P(k=4) = 20^4e^{-20}/4!=160000 \cdot 0.00000000206/24=0\\\\P(k=5) = 20^5e^{-20}/5!=3200000 \cdot 0.00000000206/120=0\\\\P(k=6) = 20^6e^{-20}/6!=64000000 \cdot 0.00000000206/720=0\\\\P(k=7) = 20^7e^{-20}/7!=1280000000 \cdot 0.00000000206/5040=0.001\\\\

P(k=8) = 20^8e^{-20}/8!=25600000000 \cdot 0.00000000206/40320=0.001\\\\P(k=9) = 20^9e^{-20}/9!=512000000000 \cdot 0.00000000206/362880=0.003\\\\P(k=10) = 20^{10}e^{-20}/10!=10240000000000 \cdot 0.00000000206/3628800=0.006\\\\P(k=11) = 20^{11}e^{-20}/11!=204800000000000 \cdot 0.00000000206/39916800=0.011\\\\

P(k\leq11)=\sum_0^{11} P(k

b) What is the probability that the number of drivers will exceed 23?

We can write this as:

P(k>23)=1-\sum_0^{23} P(k=x_i)=1-(P(k\leq11)+\sum_{12}^{23} P(k=x_i))

P(k=12) = 20^{12}e^{-20}/12!=8442485.238/479001600=0.018\\\\P(k=13) = 20^{13}e^{-20}/13!=168849704.75/6227020800=0.027\\\\P(k=14) = 20^{14}e^{-20}/14!=3376994095.003/87178291200=0.039\\\\P(k=15) = 20^{15}e^{-20}/15!=67539881900.067/1307674368000=0.052\\\\P(k=16) = 20^{16}e^{-20}/16!=1350797638001.33/20922789888000=0.065\\\\P(k=17) = 20^{17}e^{-20}/17!=27015952760026.7/355687428096000=0.076\\\\P(k=18) = 20^{18}e^{-20}/18!=540319055200533/6402373705728000=0.084\\\\

P(k=19) = 20^{19}e^{-20}/19!=10806381104010700/121645100408832000=0.089\\\\P(k=20) = 20^{20}e^{-20}/20!=216127622080213000/2432902008176640000=0.089\\\\P(k=21) = 20^{21}e^{-20}/21!=4322552441604270000/51090942171709400000=0.085\\\\P(k=22) = 20^{22}e^{-20}/22!=86451048832085300000/1.12400072777761E+21=0.077\\\\P(k=23) = 20^{23}e^{-20}/23!=1.72902097664171E+21/2.5852016738885E+22=0.067\\\\

P(k>23)=1-\sum_0^{23} P(k=x_i)=1-(P(k\leq11)+\sum_{12}^{23} P(k=x_i))\\\\P(k>23)=1-(0.021+0.766)=1-0.787=0.213

c) What is the probability that the number of drivers will be between 11 and 23, inclusive? What is the probability that the number of drivers will be strictly between 11 and 23?

Between 11 and 23 inclusive:

P(11\leq k\leq23)=P(x\leq23)-P(k\leq11)+P(k=11)\\\\P(11\leq k\leq23)=0.787-0.021+ 0.011=0.777

Between 11 and 23 exclusive:

P(11< k

d) What is the probability that the number of drivers will be within 2 standard deviations of the mean value?

The standard deviation is

\mu=\lambda =20\\\\\sigma=\sqrt{\lambda}=\sqrt{20}= 4.47

Then, we have to calculate the probability of between 15 and 25 drivers approximately.

P(15

P(k=16) = 20^{16}e^{-20}/16!=0.065\\\\P(k=17) = 20^{17}e^{-20}/17!=0.076\\\\P(k=18) = 20^{18}e^{-20}/18!=0.084\\\\P(k=19) = 20^{19}e^{-20}/19!=0.089\\\\P(k=20) = 20^{20}e^{-20}/20!=0.089\\\\P(k=21) = 20^{21}e^{-20}/21!=0.085\\\\P(k=22) = 20^{22}e^{-20}/22!=0.077\\\\P(k=23) = 20^{23}e^{-20}/23!=0.067\\\\P(k=24) = 20^{24}e^{-20}/24!=0.056\\\\

3 0
2 years ago
Which statements are true? Check all that apply. The equation |–x – 4| = 8 will have two solutions. The equation 3.4|0.5x – 42.1
NeX [460]

Answer:

1. The equation |-x -4| = 8 will have two solutions. (True)

5. The equation |0.5x – 0.75| + 4.6 = 0.25 will have no solutions. (True)

Step-by-step explanation:

1. The equation |-x -4| = 8 will have two solutions. (True)

|-x - 4| = 8

-x -4 = ±8

-x -4 = 8 and -x -4 = -8

-x = 8 + 4 and -x = -8 + 4

-x = 12 and -x = -4

x = -12 and x = 4

Therefore, it has two solutions x ∈ {-12, 4}

2. The equation 3.4|0.5x - 42.1| = -20.6 will have one solution. (False)

Since the right side of the equation has a negative value therefore, according to rules of absolute value equations, it has no solution.

3. The equation |1/2x - 3/4| = 0 will have no solutions. (False)

|(1/2)x - 3/4| = 0

(1/2)x - 3/4 = ±0

Since ±0 is the same

(1/2)x = 3/4

x = 2*3/4

x = 3/2

Therefore, it has one solution x = 3/2

4. The equation |2x – 10| = –20 will have two solutions. (False)

Since the right side of the equation has a negative value therefore, according to rules of absolute value equations, it has no solution.

5. The equation |0.5x – 0.75| + 4.6 = 0.25 will have no solutions. (True)

|0.5x – 0.75| + 4.6 = 0.25

|0.5x – 0.75| = 0.25 - 4.6

|0.5x – 0.75| = -4.35

Since the right side of the equation has a negative value therefore, according to rules of absolute value equations, it has no solution.

6. The equation |(1/8)x - 1| = 5 will have infinitely many solutions. (False)

|(1/8)x - 1| = 5

(1/8)x - 1 = ±5

(1/8)x - 1 = 5 and (1/8)x - 1 = -5

(1/8)x = 5 + 1 and (1/8)x = -5 + 1

(1/8)x = 6 and (1/8)x = -4

x = 6*8 and x = -4*8

x = 48 and x = -32

Therefore, it has two solutions x ∈ {48, -32}

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