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3 years ago

PLEASE SOMEBODY HELP ASAP

Mathematics

1 answer:

Akimi4 [234]3 years ago

5 0

Answer:

the probability of the dice is 1/6 and the probability of the spinner is 1/4.

Step-by-step explanation:

the probability of the dice=E/S=E=1 and S=6 because it is one dice. so P=1/6.

the probability of the spinner=E/S=E=1 and S=4 because the spinner is divided into 4. so P=1/4

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Please help me with this I domt understand

zzz [600]

Answer:

its A

Step-by-step explanation:

it is linear because this is happening every day, and it is negative because 1/2 is being removed every day

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3 years ago

John painted 1 side of a door with an area of 2,880 square inches in 112 hours.

anyanavicka [17]

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3 years ago

Read 2 more answers

A pizza party had 100 slices of pizza and each person ate 2 pieces of pizza. Write an

Daniel [21]

Answer:

100/2

Step-by-step explanation:

Given information :

⇒ In total, there were 100 slices of pizza

⇒ Each person ate 2 slices

With this information, the only missing piece would be the number of people who ate the pizza.

That can found using :

⇒ Total slices / Number per person

⇒ 100/2

4 0

2 years ago

HW-02 Problem No.2.1 / 10 pas 5x1 - x2 = 1 ( 3x2 - 2x1 = -3 Solve the system of linear equations by modifying it to REF and to R

IrinaVladis [17]

Answer:

$REF= \left[\begin{array}{ccc}5&-1&1\\0&\frac{-7}{5} &\frac{-18}{5}\end{array}\right]$

$RREF=\left[\begin{array}{ccc}1&0&\frac{5}{7} \\0&1&\frac{18}{7}\end{array}\right]$

Step-by-step explanation:

The augmented matrix of the system is: $\left[\begin{array}{ccc}5&-1&1\\3&-2&-3\end{array}\right]$

First we find the stepped form of A (REF):

1. We subtract 3/5 from row 1 to row 2 (R2- $\frac{3}{5}$ R1) and get the matrix

$\left[\begin{array}{ccc}5&-1&1\\0&\frac{-7}{5} &\frac{-18}{5}\end{array}\right]$

Note that this matrix is in echelon form.

Now we find the reduced row echelon form of the augmented matrix (RREF)

2. From the previous matrix, we multiply the first row by 1/5 and the second row by -5/7 and obtain the matrix:

$\left[\begin{array}{ccc}1&\frac{-1}{5} &\frac{1}{5} \\0&1&\frac{18}{7}\end{array}\right]$

3. From the previous matrix, to row 1 we add 1/5 of row 2 (R1 + $\frac{1}{5}$ R2) and we obtain the matrix

$\left[\begin{array}{ccc}1&0&\frac{5}{7} \\0&1&\frac{18}{7}\end{array}\right]$

which is the reduced row echelon form of the augmented matrix.

4 0

3 years ago

how many terms do you have to compute in order for your approximation (your partial sum) to be within 0.0000001 from the converg

Solnce55 [7]

Answer:

The answer is "4 terms".

Step-by-step explanation:

$\sum_{n=1}^{\infty} \frac{(-1)^{n+2} (0.4)^{2n+1}}{(2n+1)!} \ \ \ \ \ error \leq 0.0000001\\\\\\\sum_{n=1}^{\infty} \frac{(-1)^{n+2} (0.4)^{2n+1}}{(2n+1)!} = \frac{-1^{0+2}}{1!}(0.4) + \frac{-1^{1+2}}{3!}(0.4)^3 +\frac{-1^{2+2}}{5!}(0.4)^5+\frac{-1^{3+2}}{7!}(0.4)^7+..\\\\$

$=0.4 - \frac{0.4^3}{3!} + \frac{0.4^5}{5!} - \frac{0.4^7}{7!}+ \frac{0.4^9}{9!} - \frac{0.4^{11}}{11!}+............\\$

Note that, alternatively, positive and negative terms are given in the sequence. It is also alternating

Apply alternative test series:

$\sum_{n=1}^{\infty} (-1)^{n-1} b_n = b_1-b_2+b_3-b_4+b_5-......+(-1)^{n-1} b_n >0\\\\i) b_{n+1}$

use alternating test method we get:

$\sum_{n=1}^{\infty} \frac{-1^{n-1}}{n^2}\\\\s =0.4 - \frac{0.4^3}{3!} + \frac{0.4^5}{5!} - \frac{0.4^7}{7!}+ \frac{0.4^9}{9!} - \frac{0.4^{11}}{11!}+..\\\\b_5= \frac{0.4^9}{9!} = 0.00000000072 < \frac{1}{10^7} =0.0000001$

that's why its value is 4 terms

4 0

3 years ago