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2 years ago

Please helpppp its due nowwwww

Mathematics

1 answer:

IceJOKER [234]2 years ago

4 0

Answer:

Step-by-step explanation:

instead of (2x - 60 ) there should be (60 - 2x) because since they are parallel lines corresponding sides must be equal . It is contradictory because the size of the corresponding angles is different.

Note : Corresponding angles are the angles which are formed in matching corners or corresponding corners with the transversal when two parallel lines are intersected by any other line (i.e. the transversal).

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PLEASE HELP ME!!!!

chubhunter [2.5K]

We know, Point slope form is written as:
y - y₁ = m(x - x₁)

First, We need to Calculate the slope:
m = (y₂ - y₁) / (x₂ - x₁)
m = (2 - 1) / (-3-2)
m = 1/-5
m = -1/5

Take any one of them: Let's take (-3, 2) as our coordinate:
y - 2 = m(x - (-3))
y - 2 = -1/5(x + 3)

In short, Your Answer would be Option D

Hope this helps!

4 0

3 years ago

Find the value of n, correct to 4 significant figures.

Law Incorporation [45]

Answer:

6.708

Step-by-step explanation:

Y varies directly as the square of x (Given)

$\implies y\:\alpha \:x^2$ ......(1)

Y varies inversely as the square root of z (Given)
$\implies y\:\alpha\: \frac{1}{\sqrt z}$ ......(2)

Combining (1) & (2), we find:

$y\:\alpha \:\frac{x^2}{\sqrt z}$

$\implies y=\frac{kx^2}{\sqrt z}$ (Where k is constant of proportionality).....(3)

Now, when y = 2, x = 3 and z = 4, we find the value of k i.e. constant.

$2=\frac{k(3)^2}{\sqrt 4}$

$\implies 2=\frac{k(9)}{2}$

$\implies k =\frac{4}{9}$

Plugging the value of k in (3), we find:

$y=\frac{4x^2}{9\sqrt z}$ ....(4)

Next, in equation (4), plug y = 5, x = n and z = 16 and obtain the value of n by solving it.

$5=\frac{4(n)^2}{9\sqrt {16}}$

$\implies 5=\frac{4(n)^2}{9(4)}$

$\implies 5=\frac{(n)^2}{9}$

$\implies 5(9)=(n)^2$

$\implies 45=(n)^2$

$\implies n=\sqrt{45}$

$\implies n=6.70820393$

$\implies n\approx 6.708$

7 0

1 year ago

To find how high school students feel about hot lunches, Adam walks to the nearest high school and gives a survey postcard to ev

Eddi Din [679]

The surveyor selects the twentieth student who enters the building. The student do not enter the building to be surveyed but for other reasons and the researcher take advantage of that and selects them. This is a form of continence sampling.
The researcher selects every twentieth student and not any other. Therefore, the sampling is systematic in nature.
Finally, students selected to participate in the survey are free to choose if to participate in he survey or not. Therefore, it is voluntary kind of sampling.

Therefore, the applicable sampling methods are convenience, systematic, and voluntary.

3 0

3 years ago

Read 2 more answers

The distribution of weights for newborn babies is approximately normally distributed with a mean of 7.4 pounds and a standard de

blsea [12.9K]

Answer:

1. 15.87%

2. 6 pounds and 8.8 pounds.

3. 2.28%

4. 50% of newborn babies weigh more than 7.4 pounds.

5. 84%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 7.4 pounds

Standard Deviation, σ = 0.7 pounds

We are given that the distribution of weights for newborn babies is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

1.Percent of newborn babies weigh more than 8.1 pounds

P(x > 8.1)

$P( x > 8.1) = P( z > \displaystyle\frac{8.1 - 7.4}{0.7}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,

$P(x > 8.1) = 1 - 0.8413 = 0.1587 = 15.87\%$

15.87% of newborn babies weigh more than 8.1 pounds.

2.The middle 95% of newborn babies weight

Empirical Formula:

Almost all the data lies within three standard deviation from the mean for a normally distributed data.
About 68% of data lies within one standard deviation from the mean.
About 95% of data lies within two standard deviations of the mean.
About 99.7% of data lies within three standard deviation of the mean.

Thus, from empirical formula 95% of newborn babies will lie between

$\mu-2\sigma= 7.4-2(0.7) = 6\\\mu+2\sigma= 7.4+2(0.7)=8.8$

95% of newborn babies will lie between 6 pounds and 8.8 pounds.

3. Percent of newborn babies weigh less than 6 pounds

P(x < 6)

$P( x < 6) = P( z > \displaystyle\frac{6 - 7.4}{0.7}) = P(z < -2)$

Calculation the value from standard normal z table, we have,

$P(x < 6) =0.0228 = 2.28\%$

2.28% of newborn babies weigh less than 6 pounds.

4. 50% of newborn babies weigh more than pounds.

The normal distribution is symmetrical about mean. That is the mean value divide the data in exactly two parts.

Thus, approximately 50% of newborn babies weigh more than 7.4 pounds.

5. Percent of newborn babies weigh between 6.7 and 9.5 pounds

$P(6.7 \leq x \leq 9.5)\\\\ = P(\displaystyle\frac{6.7 - 7.4}{0.7} \leq z \leq \displaystyle\frac{9.5-7.4}{0.7})\\\\ = P(-1 \leq z \leq 3)\\\\= P(z \leq 3) - P(z < -1)\\= 0.9987 -0.1587= 0.84 = 84\%$