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Greeley [361]
2 years ago
10

What is the value of x in the equation 1/5x-2/3y = 30, when y = 15?

Mathematics
2 answers:
Nataly [62]2 years ago
8 0

Answer:

x = 200

Step-by-step explanation:

Given equation:

\sf \dfrac{1}{5}x-\dfrac{2}{3}y=30

<u><em>Steps:</em></u>

1. Substitute 15 as the value of y in the equation:

\sf \dfrac{1}{5}x-\dfrac{2}{3}(15)=30\ \textsf{[ multiply ]}\\\\\Rightarrow \dfrac{1}{5}x-\dfrac{30}{3}=30\ \textsf{[ simplify ]}\\\\\Rightarrow \dfrac{1}{5}x-10=30

2. Add 10 to both sides:

\sf \dfrac{1}{5}x-10+10=30+10\\\\\Rightarrow \dfrac{1}{5}x=40

3. Multiply both sides by 5 to isolate x:

\sf 5\left(\dfrac{1}{5}\right)x=5(40)\\\\\Rightarrow x=200

4. Check your work:

\sf \dfrac{1}{5}x-\dfrac{2}{3}y=30\ \textsf{[ substitute 200 for x, and 15 for y ]}\\\\\dfrac{1}{5}(200)-\dfrac{2}{3}(15)=30\ \textsf{[ multiply ]}\\\\\dfrac{200}{5}-\dfrac{30}{3}=30\ \textsf{[ divide ]}\\\\40-10=30\ \textsf{[ subtract ]}\\\\30=30\ \checkmark

Learn more here:

brainly.com/question/12965239

Alika [10]2 years ago
3 0

Hey there!

  • Answer :

\boxed{\bold{\green{ x = 200}}}

\\

  • Explanation:

<u>Given </u><u>expression </u><u>:</u><u> </u>

\sf{ \frac{1}{5} \green{x} -  \frac{2}{3} \orange{y} = 30 }

1) <em>Substitue 15 for</em> \sf{\orange{y}} :

\sf{ \frac{1}{5} \green{x} -  \frac{2}{3} \orange{ \times 15} = 30 } \\  \\  \implies\sf{ \frac{1}{5} \green{x} -  \frac{30}{3}  = 30 }  \\  \\  \implies\sf{ \frac{1}{5} \green{x} -  10= 30 }

\\

2) Solve for x :

\sf{ \frac{1}{5} \green{x} -  10= 30 }

⇢<u>Add</u> 10 <u>to both sides of the equation</u> :

\sf{ \frac{1}{5} \green{x} -  10  \:  \bold{+  \: 10}= 30   \: \bold{+   \: 10 }}  \\  \\   \implies \sf{\frac{1}{5} \green{x} \:  = 40 } \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

⇢<u>Multiply both </u><u>sides</u><u> of the equation</u> by 5:

\Big( \dfrac{1}{5}  \green{x}\Big)  \: \bold{ \times  \: 5}=40  \: \bold{ \times \:  5} \\  \\  \implies  \sf{\dfrac{5}{5}  \green{x} \:  = 200} \\  \\   \implies   \green{ \boxed{\sf{x = 200}}}

\\

3) Let's check our answer by replacing \green{x} with 200 and \orange{y} with 15 :

\sf{\dfrac{1}{5}\overbrace{\green{\times \: 200}}^{\green{x}} - \dfrac{2}{3}\underbrace{\orange{\times \:15}}_{\orange{y}}= \dfrac{200}{5}  -   \dfrac{30}{30}  = 40 - 10 = \boxed{ 30} }

\\ \\

▪️Learn more about first-degree equations:

↣brainly.com/question/15154296

↣brainly.com/question/14700809

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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}}}}

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The two sample t-statistic ≈ 3.43

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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.

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