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3 years ago

What number is 264 greater than 682

Mathematics

2 answers:

zmey [24]3 years ago

7 0

Greater than means to add
682 + 264 = 946

Anna007 [38]3 years ago

5 0

926 is greater than 682

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A car insurance company has determined that 9% of all drivers were involved in a car accident last year. Among the 10 drivers li

____ [38]

Answer:

The correct answer: 0.05404

Step-by-step explanation:

Given:

Binomial distribution = x

n = 10

and p = 0.09

solution:

P(X=x) =1 0Cx*(0.09^x)*((1-0.09)^(10-x)) for x=0,1,2,...,10

So the probability is calculated by the Formula:

P(X>=3) = 1-P(X=0)-P(X=1)-P(X=2)

putting the given values in the formula

= 1-10C0*(0.09^0)*((1-0.09)^(10-0))-...-10C2*(0.09^2)*((1-0.09)^(10-2))

= 0.0540400

Thus, the correct answer: 0.05404

5 0

3 years ago

Use the subtraction method to solve the system of equations. Choose the correct ordered pair. 2x-y=6

Ber [7]

Answer:

Check for (3, 0)

Put x = 3 and y = 0 in equation 2x – y = 6

2(3) – (0) = 6

6 = 6

<h2>True statement. </h2>

(3,0) is a solution of 2x – y = 6.

(ii) Check for (0, 6)

Put x = 0 and y = 6 in 2x – y = 6

2 x 0 – 6 = 6-6 = 6

<h2>False statement. </h2>

⇒ (0, 6) is not a solution of 2x – y = 6.

(iii) Check for (2, -2)

Put x = 0 and y = 6 in 2x – y = 6

2 x 2 – (-2) = 6

4 + 2 = 6

6 = 6

<h2>True statement. </h2>

⇒ (2,-2) is a solution of 2x – y = 6.

(iv) Check for (√3, 0)

Put x = √3 and y = 0 in 2x – y = 6

2 x √3 – 0 = 6

2√3 = 6

<h2>False statement. </h2>

⇒(√3, 0) is not a solution of 2x – y = 6.

(v) Check for (1/2, -5)

Put x = 1/2 and y = -5 in 2x – y = 6

2 x (1/2) – (-5) = 6

1 + 5 = 6

6 = 6

<h2>True statement. </h2>

⇒ (1/2, -5) is a solution of 2x – y = 6.

Step-by-step explanation:

Hope it is helpful......

8 0

3 years ago

I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be

Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

$P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}$

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: $g'(-1)\,(1)$ since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: $g'(-1)= 7$ as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1