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Ket [755]
3 years ago
6

Square root of 4×-1=5

Mathematics
1 answer:
borishaifa [10]3 years ago
5 0

Answer:3/2

how i did this

​ Add 11 to both sides.

4x=5+1

Simplify 5+15+1 to 66.

4x=6

Divide both sides by 4

       x=   6/4

Simplify 6/4 to 3/2

       Awnser

x=3/2

Sorry if not helpful just trying my best :)

​​  

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The two legs (labeled x) of the right triangle below have equal length. If the hypotenuse has length 5√2 , solve for x.
leonid [27]

Answer:

x=5

Step-by-step explanation:

Other than using the plain special aspect of a 45-45-90 triangle where the legs are x, x, and x√2, you can solve for this.

Since the two legs have equal length, they are both x. Using the pythagorean theorem:

(x^2)+(x^2)=50 (Because 5 squared is 25 and √2 squared is 2, multiplying them gives you 50).

You can add (x^2) and (x^2) because they are the same terms (x squared).

Simplifying like so gives you:

2x^2=50

Dividing by two on both sides:

x^2=25

Taking the square root of both sides:

x=5

6 0
3 years ago
I need the area of a rectangle, the middle is 5ft the right is 6ft and the bottom is 5ft. thank u
Scrat [10]

Answer:

length times width for area so its 5x6=30

5 0
3 years ago
35y + 14y =<br> please i need help with this asap.
dimaraw [331]

Answer: 35y + 14y = 49y

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Pls help asap!(this question is worth 20 points!)picture down below says question
diamong [38]

Answer:

138

Step-by-step explanation:

c=5

d=4

6c^2-5d+8

substitute c & d

6(5)^2-5(4)+8

6(25)-20+8

150-20+8 = 138

3 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
3 years ago
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