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miss Akunina [59]
3 years ago
14

At a pumpkin patch, if Armando guesses the weight of his pumpkin within 0.3 pounds, he gets to take the pumpkin home for free. I

f his pumpkin weighs 4.9 pounds, which two equations can be used to find the minimum and maximum weights he can guess in order to get his pumpkin for free?
x – 0.3 = 4.9 and x – 0.3 = –4.9
x + 0.3 = 4.9 and x + 0.3 = –4.9
x + 4.9 = 0.3 and x + 4.9 = –0.3
x – 4.9 = 0.3 and x – 4.9 = –0.3
Mathematics
2 answers:
Reil [10]3 years ago
7 0

Answer: last option

Step-by-step explanation:

You know that the weight of his pumpkin 4.9 pounds and that, if he guesses its weight within 0.3 pounds, he will get the pumpkin for free.

Then, to find the minimum weight he can guess  in order to get his pumpkin for free, it's necessary to write the expression:

x-0.3=4.9

Rewriting it:

x-4.9=0.3

To find the maximum weight he can guess  in order to get his pumpkin for free, it's necessary to write the expression:

x+0.3=4.9

Rewriting it:

x-4.9=-0.3

guapka [62]3 years ago
5 0

Answer:

d. x – 4.9 = 0.3 and x – 4.9 = –0.3

Step-by-step explanation: just took the test

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almond37 [142]

Answer:

What percent of the school are girls? 55%

What percent of the school are boys? 45%

How many boys are there? 162

How many girls are there? 198

Step-by-step explanation:

The number of girls = 198

The percent of girls = 198/360 x 100% = 55%

The number of boys = 360-198 = 162.

The percent of boys = 162/360 x 100% = 45%.

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Tems11 [23]
The degree is 8 ( the value of the highest exponent).
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4 years ago
In an assembly-line production of industrial robots, gearbox assemblies can be installed in one minute each if holes have been p
Illusion [34]

Answer:

The right answer is:

(a) 0.1456

(b) 18.125, 69.1202, 8.3139

Step-by-step explanation:

Given:

N = 24

n = 5

r = 7

The improperly drilled gearboxes "X".

then,

⇒ P(X) = \frac{\binom{7}{x} \binom {17}{5-x}}{\binom{24}{5}}

(a)

P (all gearboxes fit properly) = P(x=0)

                                               = \frac{\binom{7}{0} \binom{17}{5}}{\binom{24}{5}}

                                               = 0.1456

(b)

According to the question,

X = 91+5

Mean will be:

⇒ \mu = E(x)

       =E(91+5)

       =9E(1)+5

       =9.\frac{nr}{N}+5

       =9.\frac{5.7}{24} +5

       =18.125

Variance will be:

⇒ \sigma^2=Var(X)

         =V(9Y+5)

         =81.V(Y)

         =81.n.\frac{r}{N}.\frac{N-r}{N}.\frac{N-n}{N-1}

         =81.5.\frac{7}{24}.\frac{24-7}{24}.\frac{24-5}{24-1}

         =69.1202

Standard deviation will be:

⇒ \sigma = \sqrt{69.1202}

       =8.3139      

8 0
4 years ago
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viva [34]
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3 years ago
You always need some time to get up after the alarm has rung. You get up from 10 to 20 minutes later, with any time in that inte
Mama L [17]

Answer:

a) P(x<5)=0.

b) E(X)=15.

c) P(8<x<13)=0.3.

d) P=0.216.

e) P=1.

Step-by-step explanation:

We have the function:

f(x)=\left \{ {{\frac{1}{10},\, \, \, 10\leq x\leq 20 } \atop {0, \, \, \, \, \, \,  otherwise }} \right.

a)  We calculate  the probability that you need less than 5 minutes to get up:

P(x

Therefore, the probability is P(x<5)=0.

b) It takes us between 10 and 20 minutes to get up. The expected value is to get up in 15 minutes.

E(X)=15.

c) We calculate  the probability that you will need between 8 and 13 minutes:

P(8\leq x\leq 13)=P(10\leqx\leq 13)\\\\P(8\leq x\leq 13)=\int_{10}^{13} f(x)\, dx\\\\P(8\leq x\leq 13)=\int_{10}^{13} \frac{1}{10} \, dx\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot [x]_{10}^{13}\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot (13-10)\\\\P(8\leq x\leq 13)=\frac{3}{10}\\\\P(8\leq x\leq 13)=0.3

Therefore, the probability is P(8<x<13)=0.3.

d)  We calculate the probability that you will be late to each of the 9:30am classes next week:

P(x>14)=\int_{14}^{20} f(x)\, dx\\\\P(x>14)=\int_{14}^{20} \frac{1}{10} \, dx\\\\P(x>14)=\frac{1}{10} [x]_{14}^{20}\\\\P(x>14)=\frac{6}{10}\\\\P(x>14)=0.6

You have 9:30am classes three times a week.  So, we get:

P=0.6^3=0.216

Therefore, the probability is P=0.216.

e)  We calculate the probability that you are late to at least one 9am class next week:

P(x>9.5)=\int_{10}^{20} f(x)\, dx\\\\P(x>9.5)=\int_{10}^{20} \frac{1}{10} \, dx\\\\P(x>9.5)=\frac{1}{10} [x]_{10}^{20}\\\\P(x>9.5)=1

Therefore, the probability is P=1.

3 0
3 years ago
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