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3 years ago

Answer please??????????

Mathematics

2 answers:

garik1379 [7]3 years ago

7 0

W=-13

I hope this helps :)

Step2247 [10]3 years ago

7 0

Delete my answer please, I ment to answer another one.

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How do you write 23.75 as a mixed number

creativ13 [48]

23 3/4. because if it was 0.25, then you know that it is a quarter, 0.50 is half, therefore 0.75 is three quarters

8 0

3 years ago

Read 2 more answers

Let f(x) = 5x3 + 3x2 – 3 and g(x) = x2 + X - 5, fin 4) f(x) + g(x)

aleksley [76]

Answer:

$f(x)+g(x)=5x^{3} +4x^{2}+x-8$

Step-by-step explanation:

Given:

Two functions are given:

$f(x)=5x^{3} +3x^{2} -3$

$g(x)=x^{2} +x-5$

Find $f(x)+g(x)$

Solution:

$f(x)+g(x)=(5x^{3} +3x^{2} -3)+(x^{2} +x-5)$

$f(x)+g(x)=5x^{3} +3x^{2} -3+x^{2} +x-5$

$f(x)+g(x)=5x^{3} +(3x^{2} +x^{2})+x-3-5$

$f(x)+g(x)=5x^{3} +4x^{2}+x-8$

Therefore, the function of $f(x)+g(x)=5x^{3} +4x^{2}+x-8$

6 0

3 years ago

3/7 x 28/9 x 15/17 ... how do I use cancellation? my answer was 15/17 but thats not the right answer, can someone explain this a

Ne4ueva [31]

First you cancel 7 and 28, and 3 and 9, so that leaves you with 4/3 x 15/17, then you cancel 3 and 15 so you get 4x5/17 which should give you 20/17!

4 0

3 years ago

An honest die is rolled. If the roll comes out even (2, 4, or 6), you will win $1; if the roll comes out odd (1,3, or 5), you wi

jenyasd209 [6]

Answer:

(a) 50%

(b) 47.5%

Step-by-step explanation:

According to the honest coin principle, if the random variable X denotes the number of heads in n tosses of an honest coin (n ≥ 30), then X has an approximately normal distribution with mean, $\mu=\frac{n}{2}$ and standard deviation, $\sigma=\frac{\sqrt{n}}{2}$ .

Here the number of tosses is, n = 2500.

Since n is too large, i.e. n = 2500 > 30, the random variable X follows a normal distribution.

The mean and standard deviation are:

$\mu=\frac{n}{2}=\frac{2500}{2}=1250\\\\\sigma=\frac{\sqrt{n}}{2}=\frac{\sqrt{2500}}{2}=25$

(a)

To not lose any money the even rolls has to be 1250 or more.

Since, μ = 1250 it implies that the 50th percentile is also 1250.

Thus, the probability that by the end of the evening you will not have lost any money is 50%.

(b)

If the number of "even rolls" is 1250, it implies that the percentile of 1250 is 50th.

Then for number of "even rolls" as 1300,

1300 = 1250 + 2 × 25

= μ + 2σ

Then P (μ + 2σ) for a normally distributed data is 0.975.

⇒ 1300 is at the 97.5th percentile.

Then the area between 1250 and 1300 is:

Area = 97.5% - 50%

= 47.5%

Thus, the probability that the number of "even rolls" will fall between 1250 and 1300 is 47.5%.

(c)

To win $100 or more the number of even rolls has to at least 1300.

From part (b) we now 1300 is the 97.5th percentile.

Then the probability that you will win $100 or more is:

P (Win $100 or more) = 100% - 97.5%

= 2.5%.

Thus, the probability that you will win $100 or more is 2.5%.

7 0

3 years ago

Thank you so much. (20 characters required)

padilas [110]

#3.
√18²-5.7² Second Choice

#4
l²=5²+4² = 25+16 = = 41
l=√41
l= 6.4

Second Choice

#5 Only A --- First choice

#7 6 ----- Third choice

#8 V= $\pi r^2(h)= \pi 10^2(3) = 314(3)=$ 942 units³ ---Second choice

6 0

3 years ago