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2 years ago

3x-5y=-35 in slope intercept form

Mathematics

2 answers:

egoroff_w [7]2 years ago

7 0

Answer:

Step-by-step explanation:

Y=3/5x + 7

harkovskaia [24]2 years ago

7 0

The answer for this question is Y=3/5x+7

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The ratio of boys to girls in the school choir is 4:7. If there are 16 boys, how many girls are in the choir?

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2 years ago

Use the Normal model N(1133,78) for the weights of steers.

Rudiy27

Answer:

Step-by-step explanation:

Let $X\sim N(1133, 78)$ .

Here, the mean is 1133 and standard deviation is 78.

Since we don't have the table for N(1133, 78), so we use the standard table for N(0,1),

All we have to do is find, within the table the specific percentile.

now to find (42nd percentile) 0.42 in the normal table N(0,1) and extract the number on the row and column.

$P(X\leq x)=0.42$

Using: $Z=\frac{X-\mu}{\sigma}$ where $\mu$ is the mean and $\sigma$ is the standard deviation.

$P(\frac{X-1133}{78} \leq \frac{x-1133}{78})=0.42$

let $z=\frac{x-1133}{78}$

then, we have

$P(Z\leq z)=0.42$

Now, using Standard Normal table to find the value of z- score;

Usually, tables are set up as the probability that a number, z is less than or equal to Z.

i.e, $z= -0.20$

Now, putting this value in $z=\frac{x-1133}{78}$ , to find x;

$\frac{x-1133}{78}=-0.20$

On Simplify, we get;

$x=1,117.4$ pounds

Similarly, find the weight for 91st percentile.

Follow the same steps that we have done in part (a),

$P(X\leq x)=0.91$ or

$P(\frac{X-1133}{78} \leq \frac{x-1133}{78})=0.91$

⇒ $P(Z \leq \frac{x-1133}{78})=0.91$

let $z=\frac{x-1133}{78}$

then, we have

$P(Z\leq z)=0.91$

Now, use normal table value to find the z-score;

Usually, tables are set up as the probability that a number, z is less than or equal to Z

we have, z= 1.34

putting the z value in $z=\frac{x-1133}{78}$ we get,

$\frac{x-1133}{78}=1.34$

On simplify, we get

x= 1,237.52 pounds

the interquartile range (IQR)=3rd Quartile - 1st quartile

First find the 1st quartile:

$P(\frac{X-1133}{78} \leq \frac{x-1133}{78})=0.25$

⇒ $P(Z \leq \frac{x-1133}{78})=0.25$

let $z=\frac{x-1133}{78}$

then, we have

$P(Z\leq z)=0.25$

Now, we use normal table to find that z=-0.675

putting the z value in $z=\frac{x-1133}{78}$ we get,

$\frac{x-1133}{78}=-0.675$

On simplify, we get

x= 1,080.35 pounds

Similarly, for third quartile

By symmetry z=0.675

putting the z value in $z=\frac{x-1133}{78}$ we get,

$\frac{x-1133}{78}=0.675$

On simplify, we get

x= 1,185.65 pounds

then, IQR = 1185.65 -1080.35=105.3 pounds

7 0

3 years ago

WILL MARK BRAINLIEST IF CORRECT PLEASE HELP

Romashka [77]

Answer:

$\frac{4\sqrt{7}}{7} \text{ and } \frac{3\sqrt{7}}{7}$

Step-by-step explanation:

if $\sec x = \frac{4}{3}$ , then that means that it is the angle of a right triangle with the hypotenuse equal to 4 and one of the legs (specifically the adjacent leg) equal to 3, since sec(x) is defined as hypotenuse over the adjacent side.

The other leg (specifically the opposite one) will be equal to $\sqrt{16-9} = \sqrt{7}$ by the pythagorean theorem, so csc(x) = hypotenuse over opposite side = $\frac{4}{\sqrt{7}}=\frac{4\sqrt{7}}{7}$ .

cot(x) is defined as adjacent over opposite, which is equal to $\frac{3}{\sqrt{7}}=\frac{3\sqrt{7}}{7}$

That means that the first option, the leftmost one, is correct.

3 0

2 years ago