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liraira [26]
3 years ago
8

MARK BRAINLIEST AND 25 POINTS In the figure below, BCA is similar to BDC is similar to CDA.

Mathematics
1 answer:
Zinaida [17]3 years ago
8 0

Answer:

a²=c². b²=c²

a²+b²=c²+c²

a²+b²=c(c+c)

a²+b²=c(c)

a²+b²=c²

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Select all expressions that are true when x=2
melamori03 [73]

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3 years ago
Some researchers have conjectured that stem-pitting disease in peach-tree seedlings might be controlled with weed and soil treat
stiks02 [169]

Answer:

Part A

b. 14.6 ± 7.38

Part B

b. 3.43

Part C

a. P-value < 0.01

Part D

b. There is sufficient evidence to reject the null hypothesis

Step-by-step explanation:

Part A

The given data are;

The number of seedlings in the field = 20

The number of seedlings selected to receive herbicide A = 10

The number of seedlings selected to receive herbicide B = 10

The height in centimeters of seedlings treated with Herbicide A, \overline x _1 = 94.5 cm

The standard deviation, s₁ = 10 cm

The height in centimeters of seedlings treated with Herbicide B, \overline x _2 = 109.1 cm

The standard deviation, s₂ = 9 cm

The 90% confidence interval for μ₂ - μ₁, is given as follows;

\left (\bar{x}_{2}- \bar{x}_{1}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

The critical-t at 95% and n₁ + n₂ - 2 degrees of freedom is given as follows;

The degrees of freedom, df = n₁ + n₂ - 2 = 10 + 10 - 2 = 18

α = 100% - 90% = 10%

∴ For two tailed test, we have, α/2 = 10%/2 = 5% = 0.05

t_{(0.025, \, 18)} = 1.734

C.I. = \left (109.1- 94.5 \right )\pm 1.734 \times \sqrt{\dfrac{10^{2}}{10}+\dfrac{9^{2}}{10}}

C.I. ≈ 14.6 ± 7.37714603353

The 90% C.I. ≈ 14.6 ± 7.38

b. 14.6 ± 7.38

Part B

With the hypotheses are given as follows;

H₀; μ₂ - μ₁ = 0

Hₐ; μ₂ - μ₁ ≠ 0

The two sample t-statistic is given as follows;

t=\dfrac{(\bar{x}_{2}-\bar{x}_{1})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}+\dfrac{s _{2}^{2}}{n_{2}}}}

t-statistic=\dfrac{(109.1-94.5)}{\sqrt{\dfrac{10^{2} }{10}+\dfrac{9^{2}}{10}}} \approx 3.43173361147

The two sample t-statistic ≈ 3.43

b. 3.43

Part C

From the t-table, the p-value, we have, the p-value < 0.01

a. P-value < 0.01

Part D

Given that a significance level of 0.05 level is used and the p-value of 0.01 is less than the significance level, there is enough statistical evidence to reject the null hypothesis

b. There is sufficient evidence to reject the null hypothesis.

4 0
3 years ago
Someone please help with these and explain your answer for both!!! I will give brainliest if correct!!!
goldenfox [79]

Answer:

C

Step-by-step explanation:

We have congruent angles in the middle and 2 sides are congruent as well which makes (Angle-Side-Side) which doesn't work for congruent triangles.

3 0
2 years ago
Read 2 more answers
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tangare [24]

Answer:

2 notebooks

Step-by-step explanation:

I understand you already figured out that it was 3 notebooks, but when I calculated it, I believe he could only buy 2...

Personally, I would show my work by first doing the calculation: 2.75[1 notebook price]+ (3[pens/pencils per notebook]*1.10[pen/pencil cost])= $9.075 per notebook and 3 pens

If you then showed that you multiplied 9.075 by 2... you would get $18.15 which is the closest you could get to his limit of $22 without going over

If we applied your answer of 3 notebooks, then the total would be $27.225 which goes over the money limit.

Hope I helped!

6 0
3 years ago
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