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3 years ago

Chas and Tyler have been collecting video games. Chas

Mathematics

1 answer:

marshall27 [118]3 years ago

4 0

Answer:

Step-by-step explanation:

m+25=65

65-25=40

m=40

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My family and I are so confused on this! Help please!

-BARSIC- [3]

Answer:

10) Correct

11) Each row is equal to 12 when you multiply them together.

12) 24

Step-by-step explanation:

12) You multiply 4*3*2 to get the volume of 24, which is also the number of cubes needed to make the shape.

Please give brainliest if I helped! :)

3 0

2 years ago

1/2 - 3 1/4 1/2 - 4/5 1/6 + -3 3/4

Olenka [21]

Answer:

1/2 - 3 1\4= -11/4

1\2 - 4\5= -3\4

1\6 + -3 3\4= 43\12

3 0

3 years ago

Pls anwer this qestion it is worth 30 points

Sveta_85 [38]

Answer:

Step-by-step explanation:

25b² + 70b + ....

( 5b + 7)² = 5×5b² + 2×5×7b + 7×7 = 25b² + 70b + 49

6 0

2 years ago

To make cranberry jam you need 12 cups of sugar for every 16 cups of cranberries. Find the amount of sugar needed for 4 cups of

kipiarov [429]

Answer:

It would be three cups

Step-by-step explanation:

16/12=4/x

Divide 16 by 4 to get 4 cups of cranberries, and do the same to the sugar

3=x

5 0

2 years ago

The probability density function of the time you arrive at a terminal (in minutes after 8:00 A.M.) is f(x) = 0.1 exp(−0.1x) for

Blababa [14]

$f_X(x)=\begin{cases}0.1e^{-0.1x}&\text{for }x>0\\0&\text{otherwise}\end{cases}$

a. 9:00 AM is the 60 minute mark:

$f_X(60)=0.1e^{-0.1\cdot60}\approx0.000248$

b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is

$\displaystyle\int_{15}^{30}f_X(x)\,\mathrm dx\approx0.173$

c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is

$\displaystyle\int_0^{40}f_X(x)\,\mathrm dx\approx0.982$

The probability of doing so for at least 2 of 5 days is

$\displaystyle\sum_{n=2}^5\binom5n(0.982)^n(1-0.982)^{5-n}\approx1$

i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.

d. Integrate the PDF to obtain the CDF:

$F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x$

Then the desired probability is

$F_X(30)-F_X(15)\approx0.950-0.777=0.173$

7 0

3 years ago