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adelina 88 [10]
2 years ago
14

1. Which best describes your ability to solve quadratic equations?​

Mathematics
1 answer:
-BARSIC- [3]2 years ago
3 0

Answer:

you just have to do it yourself

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Please answer immediately.​
salantis [7]

1. Angle L

2. A

3. Side PL

4. Side AR

5. A and K

6. B

7. Side Br

8. A

9. T

10. TA

4 0
3 years ago
Hi how can I graph this if it dosn't have a slope?<br> 1. y = (2/3)x - 1 <br> y = -x + 4
12345 [234]

System of Linear Equations entered :

[1] y - 2x/3 = -1

[2] y + x = 4

// To remove fractions, multiply equations by their respective LCD

Multiply equation [1] by 3

// Equations now take the shape:

[1] 3y - 2x = -3

[2] y + x = 4

Graphic Representation of the Equations :

-2x + 3y = -3 x + y = 4

Solve by Substitution :

// Solve equation [2] for the variable x

[2] x = -y + 4

// Plug this in for variable x in equation [1]

[1] 3y - 2•(-y +4) = -3

[1] 5y = 5

// Solve equation [1] for the variable y

[1] 5y = 5

[1] y = 1

// By now we know this much :

y = 1

x = -y+4

// Use the y value to solve for x

x = -(1)+4 = 3

I hope this help you

8 0
3 years ago
one of the solutions to x2 − 2x − 15 = 0 is x = −3. what is the other solution? x = −5 x = −1 x = 1 x = 5
Goryan [66]
x^{2} -2x-15=0
P(-15)
s(-2)
x^{2} -5x+3x-15
=x(x-5)+3(x-5)
=(x+3)(x-5)
x=5
3 0
3 years ago
All the people of a neighborhood pooled together and won the lottery they won 10,000,000 and each person got a 0.02 share. How m
natka813 [3]

Answer:

500000000

Step-by-step explanation:

10000000 ÷ 0.02

4 0
3 years ago
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
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