Solve 6b<24 or 4b+12>4

Solve the quadratic by factoring.<br> x^2+ 15x = 5x – 24

vagabundo [1.1K]

I think it is X=-4,-6

7 0

3 years ago

The total charge that has entered a circuit element is q(t) = 6 (1 - e-7t) when t ≥ 0 and q(t) = 0 when t < 0. The current in

Bess [88]

Answer:

$A=42, B=-7$

Step-by-step explanation:

The current function of time is defined as follows:

$I(t)=\frac{dq(t)}{dt}$

where $q(t)$ is the charge function.

For the given charge function of time $q(t)=6\left( 1-e^{-7t}\right)$ we have the following current function:

$I(t)=\frac{d}{dt} \left(6\left( 1-e^{-7t}\right)\right)=42e^{-7t}$

In the problem it is proposed that $I(t)=Be^{-At}$ .

Examining the expression of $I(t)$ we obtained by deriving $q(t)$ with the expression proposed by the problem and comparing term by term:

$I(t)=Be^{-At}=42e^{-7t}$

We conclude that $A=-7$ and $B=42$ .

4 0

3 years ago

Miguel is playing a game in which a box contains four chips with numbers written on them. Two of the chips have the number 1, on

Iteru [2.4K]

Answer:

$P(X_i=2) =\dfrac{1}{6}$

$P(X_i=-1) =\dfrac{5}{6}$

Step-by-step explanation:

Given the numbers on the chips = 1, 1, 3 and 5

Miguel chooses two chips.

Condition of winning: Both the chips are same i.e. 1 and 1 are chosen.

Miguel gets $2 on winning and loses $1 on getting different numbers.

To find:

Probability of winning $2 and losing $1 respectively.

Solution:

Here, we are given 4 numbers 1, 1, 3 and 5 out of which 2 numbers are to be chosen.

This is a simple selection problem.

The total number of ways of selecting r numbers from n is given as:

$_nC_r = \frac{n!}{r!(n-r)!}$

Here, n = 4 and r = 2.

So, total number of ways = $_4C_2 = \frac{4!}{2!\times 2!} = 6$

Total number of favorable cases in winning = choosing two 1's from two 1's i.e. $_2C_2 = \frac{2!}{2! 0! } = 1$

Now, let us have a look at the formula of probability of an event E:

$P(E) = \dfrac{\text{Number of favorable ways}}{\text{Total number of ways}}$

So, the probability of winning.

$P(X_i=2) =\dfrac{1}{6}$

Total number of favorable cases for -1: (6-1) = 5

So, probability of getting -1:

$P(X_i=-1) =\dfrac{5}{6}$

Please refer to the attached image for answer table.

7 0

3 years ago

Which value(s) from the set {5,7,9,11,13} make the inequality w-4<8 true?

ale4655 [162]

The answer set is {5,7,9,11}

8 0

3 years ago

Read 2 more answers

The researcher is deciding between a 95% confidence level and a 99% confidence level. Compared to a 95% confidence interval, a 9

sineoko [7]

Answer:

Option (a) narrower and would involve a larger risk of being incorrect.

Step-by-step explanation:

We have to compare a 95% confidence interval to a 99% confidence interval.

For a 95% confidence interval and 99% confidence interval, confidence level is lower for 95% confidence interval.
With a lower confidence level, there would be less margin of error.
Thus, 95% confidence interval is narrower as compared to the 99% confidence interval.
Since, a 95% interval is narrower there are less values included in the interval that could give right estimate for the parameter.

Thus, we can write,

Compared to a 95% confidence interval, a 99% confidence interval will be Option (a) narrower and would involve a larger risk of being incorrect.

5 0

3 years ago