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3 years ago

For what values of X are the statements below true

Mathematics

1 answer:

Galina-37 [17]3 years ago

3 0

Answer:

See solutions below

Step-by-step explanation:

For what values of X are the statements below true

A. 1x>x+1

x >x+1

x-x>1

0x > 1

x > 1/0

X >∞

B) |1-x|>3

The fucntion can both be positive and negative

For the negative function

-(1-x) > 3

-1+x > 3

x > 3+1

x > 4

For the positive function

1-x > 3

-x > 3 - 1

-x > 2

x < -2

Hence the required solutions are x > 4 and x < -2

c) For the equation

|x-15| < 0

-(x - 15) < 0

-x + 15 < 0

-x < -15

x > 15

Also x-15 < 0

x < 0+15

x < 15

Hence the required solution is x > 15 and x < 15

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In the flowering plant, white flowers (B) are dominant over red flowers (b), and short plants (E) are dominant over tall flower

lutik1710 [3]

<u>Question Completion</u>

(a)What is your null hypothesis?

(b)What is your expected phenotypic ratio based on Mendelian inheritance?

(c)Calculate the expected number of flowers you should have gotten based on the Mendelian inheritance. Then calculate a chi-square value, degrees of freedom, and a p-value.

Chi-square statistic: _____
Degrees of freedom (# phenotypes -1):
P-value:

(d)Interpret your results. Do you reject it or fail to reject your null hypothesis (please restate the null)?

Answer:

(a) $H_0:$The given data fit the predicted phenotype$

(b)9:3:3:1

(c)

Chi-square statistic: 3.8914
Degrees of freedom (# phenotypes -1) =3
P-value: 0.2734

(d)We fail to reject the null hypothesis.

Step-by-step explanation:

In the flowering plant, white flowers (B) are dominant over red flowers (b), and short plants (E) are dominant over tall flowers (e). An F2 generation was created by crossing two F1 individuals (each BbEe).

(a)The null hypothesis is:

$H_0:$The given data fit the predicted phenotype$

(b)The gametes are BE, Be, bE and be.

The offsprings are presented in the table below:

$\left|\begin{array}{c|cccc}&BE&Be&bE&be\\--&--&--&--&--\\BE&BE&BE&BE&BE\\Be&BE&Be&BE&Be\\bE&BE&BE&bE&bE\\be&BE&Be&bE&be\end{array}\right|$

The expected phenotypic ratio based on Mendelian inheritance

BE:Be:bE:be=9:3:3:1

(c)

$\left|\begin{array}{c|c|c|c|c|c}$Phenotype&Observed&$Expected&O-E&(O-E)^2&\dfrac{(O-E)^2}{E} \\-----&--&--&--&--&--\\$White short(BE)&206&\frac{9}{16}*404 \approx 227 &-21&441&1.9427\\$Red, short(bE)&83&\frac{3}{16}*404 \approx 78 &5&25&0.3205\\$White, tall(Be)&85&\frac{3}{16}*404 \approx 78 &7&49&0.6282\\$Red, tall(be)&30&\frac{1}{16}*404 \approx 25 &5&25&1\\-----&--&--&--&--&--\\$Total&404&--&--&--&3.8914\end{array}\right|$

Therefore:

Chi-square statistic: 3.8914
Degrees of freedom (# phenotypes -1): 4-1 =3
P-value: 0.2734

(d) Our null hypothesis is:

$H_0:$The given data fit the predicted phenotype$

Since p>0.05, the given data fit the predicted phenotypic ratio.

We, therefore, fail to reject the null hypothesis.

The difference in the observed and expected are sosmall that they can be attributed to random chance.

8 0

3 years ago

I need help plz !!!!

svetlana [45]

X = 66 or 180 - 114, which = 66

3 0

4 years ago

The manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .

Misha Larkins [42]

Answer:

0.1507 or 15.07%.

Step-by-step explanation:

We have been given that the manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .016 millimeters. To be acceptable the diameter needs to be between 21.97 and 22.03 millimeters.

First of all, we will find z-scores for data points using z-score formula.

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

z = z-score,

x = Sample score,

$\mu$ = Mean,

$\sigma$ = Standard deviation.

$z=\frac{21.97-22}{0.016}$

$z=\frac{-0.03}{0.016}$

$z=-0.1875$

Let us find z-score of data point 22.03.

$z=\frac{22.03-22}{0.016}$

$z=\frac{0.03}{0.016}$

$z=0.1875$

Using probability formula $P(a$ , we will get:

$P(-0.1875$

Therefore, the probability that a randomly selected ball bearing will be acceptable is 0.1507 or 15.07%.

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4 years ago

Cody said that dividing a unit fraction by a whole number is the same as multiplying the unit fractions by a unit fraction with

LekaFEV [45]

Answer:

True

Step-by-step explanation:

Hi, the statement is true, because to divide a fraction we have to turn the second fraction upside down and multiply them. Since a whole number x can be written as x/1, it also applies to the case of dividing a fraction by a whole number.

We can prove it with an example.

Dividing a unit fraction by a whole number

1/2 ÷ 2 = (1x1) / (2x2) = 1/4

Multiplying the unit fractions by a unit fraction with a whole number as a denominator

1/2 x 1/2= (1x1) / (2x2) = 1/4

4 0

3 years ago

Read 2 more answers

A marble is randomly selected from a bag. The probability of selecting a marble with dots on it is 0.2. The probability of selec

marin [14]

Answer:

0.5

Step-by-step explanation:

Let D be the event of selecting a marble with dots.

Let P be the event of selecting a purple marble.

The probability of selecting a marble with dots, P(D)=0.2

The probability of selecting a marble that is both purple and has dots, $P(D \cap P)=0.1$

We want to determine the probability of selecting a purple marble given that the marble has dots on it, P(P|D)

By the definition of conditional probability:

$P(P|D)= \dfrac{P(P \cap D)}{P(D)} \\= \dfrac{0.1}{0.2}\\ =0.5$

The probability of selecting a purple marble given that the marble has dots on it is 0.5.

4 0

3 years ago