All categories

3 years ago

Help with all please help!!!!! 11 points!!!!

Mathematics

1 answer:

Paladinen [302]3 years ago

5 0

1. You’d use a number line because since the second number is a negative it’s subtracted from six so you would count 8 don’t the number line from 6

2.-16+8=-8

3.gained 4 yrds I think

You might be interested in

U divided by 25 =13 please do step by step and show work thanks :))

Anna71 [15]

Answer U/25=13
U=13 divided 1/25
U=13 multiply 25
U=325

6 0

3 years ago

Fiona has met her goal of exercising 150 minutes a week. She has kept it up for six months and is sure that it is now solidly pa

cricket20 [7]

Answer:

Step-by-step explanation:

You are always able to do better, and Fiona needs to reflect

4 0

2 years ago

Find maclaurin series

Mumz [18]

Recall the Maclaurin expansion for cos(x), valid for all real x :

$\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

$\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}$

The first 3 terms of the series are

$\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}$

and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

$\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}$

which amounts to replacing the x with √x in the expansion of cos(√5 x) :

$\cos\left(\sqrt{5x}\right) \approx 1 - \dfrac{5x}2 + \dfrac{25x^2}{24}$

7 0

2 years ago

Eric throws a biased coin 10 times. He gets 3 tails. Sue throw the same coin 50 times. She gets 20 tails. Aadi is going to throw

iren2701 [21]

Answer:

(1) The correct option is (A).

(2) The probability that Aadi will get Tails is $\frac{2}{5}$ .

Step-by-step explanation:

It is provided that:

Eric throws a biased coin 10 times. He gets 3 tails.
Sue throw the same coin 50 times. She gets 20 tails.

The probability of tail in both cases is:

$P(\text{Tail})=\frac{3}{10}$
$P(\text{Tail})=\frac{20}{50}=\frac{2}{5}$

(1)

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

In this case we need to compute the proportion of tails.

Then according to the Central limit theorem, Sue's estimate is best because she throws it <em>n = </em>50 > 30 times.

Thus, the correct option is (A).

(2)

As explained in the first part that Sue's estimate is best for getting a tail, the probability that Aadi will get Tails when he tosses the coin once is:

$P(\text{Tail})=\frac{20}{50}=\frac{2}{5}$

Thus, the probability that Aadi will get Tails is $\frac{2}{5}$ .

7 0

3 years ago

Read 2 more answers

Anybody can help me with this ?

Sedaia [141]

I was just learning that in class. It doesn’t make sense

8 0

2 years ago

Read 2 more answers