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Sav [38]
2 years ago
8

5th grade math. Correct answer will be marked brainliest.

Mathematics
2 answers:
lord [1]2 years ago
8 0
Each bag has 1 (1x12 =12) pound of raisins and you have to add two 0’s at the end
amid [387]2 years ago
7 0
Each bag has 1.25 lbs of raisins, you had to write 2 zeros at the end
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How do you figure this out? Keith plans to buy a car when he is 16. He currently has $3,000 saved up to buy the car. If his pare
kobusy [5.1K]

Answer:

He would need to save 500 dollars a month because 500x for two years (24 months) equals 12,000.

12,000 plus 3000 equals to 15,000

Step-by-step explanation:

hope this helps :)

3 0
2 years ago
A certain Bookstore that sells A Million books wants to hire you, and you get the job if you can answer these problems correctly
Maslowich

Answer:

a) P ( X = 2 ) = 0.23028

b) P ( X < 4 ) = 0.95689

c) P ( X ≥ 3 | X ≥ 2 ) = 0.38292

Step-by-step explanation:

Given:-

- The parameter for the poisson distribution is given, λ = 1.3.

- Declare a random variable (X) which is the number of books sold in the next minute:

                               X ~ Po (1.3)

Find:-

a) What's the probability that the store will sell 2 books in the next minute? b) What's the probability that the store will sell less than 4 books in the next minute? c) What's the probability that the store will sell at least 3 books in the next minute given that it sells at least 2 books in the next minute?

Solution:-

a) The required probability P ( X = 2 ). Can be computed by using the pmf for the poisson distribution:

                       P(X = x ) =\frac{ (lambda)^k e^(^-^l^a^m^b^d^a^)}{k!}\\\\P(X = x )  =\frac{ (1.3)^k e^(^-^1^.^3^)}{k!}

Where, "k" is the number of books sold in next minute.

- Now compute P ( X = 2 ) :

                       P(X = 2 )  =\frac{ (1.3)^2 e^(^-^1^.^3^)}{2!}\\\\P(X = 2 )  = 0.23028    

b) The required probability P ( X < 4 ). Can be computed by using the pmf for the poisson distribution and summing individual terms from 0 - 3:

                      P(X < 4 )  = P ( X = 0) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 )\\\\P(X < 4 )  = \frac{ (1.3)^0 e^(^-^1^.^3^)}{0!}+ \frac{ (1.3)^1 e^(^-^1^.^3^)}{1!}+ \frac{ (1.3)^2 e^(^-^1^.^3^)}{2!} + \frac{ (1.3)^3 e^(^-^1^.^3^)}{3!}\\\\P(X < 4 )  = 0.27253 + 0.35429 + 0.23028 + 0.09979 = 0.95689

c) The required probability P ( X ≥ 3 | X ≥ 2 ). We have to consider the conditional probability as follows:

                   P ( X ≥ 3 | X ≥ 2 ) = P ( X ≥ 3 & X ≥ 2 ) / P (X ≥ 2 )

                                               = P ( X ≥ 3 ) / P (X ≥ 2 )

                                               = P ( X > 2 ) / P ( X > 1 )

                                               = [ 1 - P ( X ≤ 2 ) ] / [ 1 - P ( X ≤ 1 ) ]  

- Compute P ( X ≤ 2 ) & P ( X ≤ 1 ) using pmf:

                     P ( X ≤ 2 ) = 0.27253 + 0.35429 + 0.23028

                                      = 0.8571

                     P ( X ≤ 1 ) = 0.27253 + 0.35429

                                      = 0.62682

- Use the expression developed for conditional probability to evaluate the required probability:

                     P ( X ≥ 3 | X ≥ 2 ) = [ 1 - P ( X ≤ 2 ) ] / [ 1 - P ( X ≤ 1 ) ]

                                                  = [ 1 - 0.8571 ] / [ 1 - 0.62682 ]

                                                  = 0.38292

8 0
3 years ago
Classify a transformation as a rotation, a reflection, or a translation.
krek1111 [17]

Answer:

a transformation could be a rotation, reflection, and/or a translation.

Step-by-step explanation:

In geometry, transformation refers to the movement of objects in the coordinate plane.

Therefore, a transformation could be a rotation, reflection, and/or a translation.

4 0
3 years ago
I need help with number 8 and 9 someone please help ASAP
vichka [17]
I just know #8 but it's c
7 0
3 years ago
Find the simple interest paid to the nearest cent for each principal, interest rate, and time.
igor_vitrenko [27]

<u>Answer:</u>

The simple interest for $152, 2.5%, 18 month is $5.7

<u>Solution:</u>

Given that, principal amount = $ 152, interest rate = 2.5 % and time period = 18 months.

Now we have to calculate the simple interest for above given values.

We know that, simple interest is given as

=\frac{\text { principal amount } \times \text { interest rate } \times \text { time period in years }}{100}

By substituting the given values, we get

\text { Simple interest }=\frac{152 \times 2.5 \times 18 \mathrm{months}}{100}

By converting 18 months to years we get,

\begin{array}{l}{=\frac{152 \times 2.5 \times \frac{18}{12}}{100}} \\\\ {=\frac{152 \times 2.5 \times \frac{3}{2}}{100}} \\\\ {=\frac{152 \times \frac{3}{2}}{40}} \\\\ {=\frac{76 \times 3}{40}=\frac{19 \times 3}{10}} \\\\ {=\frac{57}{10}=5.7}\end{array}

Hence, the simple interest is $5.7

3 0
3 years ago
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