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SCORPION-xisa [38]
2 years ago
15

I just need the answer please dont explain ​

Mathematics
2 answers:
timama [110]2 years ago
7 0
The answer is 12 :))))
zavuch27 [327]2 years ago
3 0
Use photomath it’s faster and you get correct Answers!<3 have a good day
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Write the following integers in a decreasing order. 1. 0,7,-3,9,-132,36​
Elan Coil [88]
ANSWER:
36 > 9 > 7 > 0 > -3 > -132

Hope it helps u!
6 0
3 years ago
Read 2 more answers
6) Determine if the two figures are similar. Present an informal argument to present
Nadusha1986 [10]

Answer:

yes they are similar

step 1: flip it across the y-axis

step 2: translate it 4 units to the right and 5 units down

step 3: dilate it by a factor of 1/2

Step-by-step explanation:

7 0
3 years ago
Multi-Step Equation 2: Solve for x. Show your steps.<br> 6 - 5x + 2 = 3x - 12 + 2x
amm1812

Answer:

The answer is x = 8/5

Step-by-step explanation:

6 - 5x + 2 = 3x - 12 + 2x

4 - 5x = 5x - 12

4 = 10x - 12

16 = 10x

8/5 = x

4 0
2 years ago
Evaluate:
Ivanshal [37]

Answer:the answer is -17

Step-by-step explanation:

6 0
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
2 years ago
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