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2 years ago

Please help me i'll mark you brainlest or whatever.

Mathematics

2 answers:

Jlenok [28]2 years ago

5 0

I can’t see the picture

nexus9112 [7]2 years ago

4 0

The measurement is 60 degrees because

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Which fraction equals the ratio of rise to run between the points (0, 0) and (6, 7)? A. B. C. D.

Svetlanka [38]

Answer:

7 / 6

Step-by-step explanation:

Given the points:

points (0, 0) and (6, 7)

Point 1 : x1 = 0 ; y1 = 0

Point 2 : x2 = 6 ; y2 = 7

The rise = y2 - y1 = 7 - 0 = 7

The run = x2 - x1 = 6 - 0 = 6

Ratio of Rise to Run = Rise / Run = 7 / 6

3 0

2 years ago

How do I solve <img src="https://tex.z-dn.net/?f=%20%20%5Csin%28%20a%20%29%20%20%3D%20.8325" id="TexFormula1" title=" \sin(

Serga [27]

Answer:

$a=56.4\degree$

Step-by-step explanation:

Let $a$ be an acute angle.

To solve: $\sin(a)=0.8325$ , we take the sine inverse of both sides to obtain:

$\sin^{-1}(\sin(a))=\sin^{-1}(0.8325)$

Recall composition property of a function and its inverse function.

$f^{-1}(f(x))=x$

We apply this property to the left hand side to obtain;

$a=\sin^{-1}(0.8325)$

We now use our scientific calculator to obtain;

$a=56.3564115\degree$

We round to the nearest tenth to obtain;

$a=56.4\degree$

4 0

2 years ago

Yearly tuition, housing, and fees for a local college are $10,500, $1300, and $2170 respectively. You have saved $3100 and have

Natalija [7]

Answer:

4,370

Step-by-step explanation:

add 10,500, 1,300, and 2,170

then add 3,100 and 6,500

subtract both sums

got it

4 0

2 years ago

A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with

Ratling [72]

Answer:

$P(X>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)$

And we can find this probability using the complement rule and with excel or the normal standard table:

$P(z>1.61)=1-P(z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(10.5,0.3)$

Where $\mu=10.5$ and $\sigma=0.3$

We are interested on this probability

$P(X>10.983)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)$

And we can find this probability using the complement rule and with excel or the normal standard table:

$P(z>1.61)=1-P(z$

8 0

3 years ago

What is the coefficient of the x^5y^5- term in the biomial expansion of (2x-3y)^10

jasenka [17]

Answer:

The coefficient of $x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552$

Solution:

The given expression is $(2 x-3 y)^{10}$

As per binomial theorem, we know,

$(x+y)^{n}=\sum n C_{k} x^{n-k} y^{k}$

Now here a = 2x, b = (- 3y) and n = 10 and k = 0,1,2,….10

Now $x^{5}\times y^5$ will be the 6 term where k =5

Now, $\mathrm{T}_{6}=10 \mathrm{C}_{5} \times(2 \mathrm{x})^{(10-5)} \times(-3 \mathrm{y})^{5}=10 \mathrm{C}_{5} 2^{5} \times \mathrm{x}^{5} \times(-3)^{5} \times \mathrm{y}^{5}$

So, the coefficient of $x^{5} \times y^{5} \text { is }=10\left(5 \times 2^{5} \times(-3)^{5}\right$ .

$10 \mathrm{C}_{5}=\frac{10 !}{5 ! \times(10-5) !}=\frac{10 !}{5 !+5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}=\left(\frac{30240}{120}\right)=252$

The coefficient of $x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552$

6 0

3 years ago