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3 years ago

Find the value of this expression if x = 2 and x=1 xy^2 / 5

Mathematics

1 answer:

natita [175]3 years ago

5 0

If x = 2 and y = 1, the value of the equation is

$\frac{xy^2}{5}$ = $\frac{2*1}{5}$ = $\frac{2}{5}$

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Katie and becca each bought a new book for $50 . Katie sold her book to the bookstore for 25% less than the original price. . Be

r-ruslan [8.4K]

$\text{Lets find the prices in which they sold their books for}\\\\\text{Katie:}\\\\\text{She sold the book for 25\% less than the original price, so multiply 50 by}\\\text{0.25 and subtract}\\\\50*0.25=12.5\\\\50-12.5=37.50\\\\\text{Becca:}\\\\\text{Becca sold the book for 40\% less than the original price, calculate how}\\\text{much she sold it for}\\\\50*0.40=20\\\\50-20=30\\\\$ $\text{We need to find how much more money Katie made than Becca, so subtract}\\\text{Becca's cost from Katie's cost}\\\\37.50-30=7.50\\\\\boxed{\text{Katie received \$7.50 more than Becca}}$

4 0

3 years ago

Read 2 more answers

The mean of a set of credit scores is Mu = 690 and Sigma = 14. Which statement must be true about z694? z694 is within 1 standar

Kitty [74]

Answer:

I believe it's B

Step-by-step explanation:

hope this helps!

5 0

2 years ago

Read 2 more answers

-7x+4= 18 set notation interval notation

denpristay [2]

Answer:

Step-by-step explanation:

-7x+4=18

-7x=14

Divide-7 on both sides of the equal sign which is going to leave you with x=-2

7 0

2 years ago

Assume that the national credit card interest rate is 12.83 percent. A study of 62 college students finds that their average int

meriva

Answer:

b) 95 percent confidence interval for this single-sample t test

[11.64, 16.36]

Step-by-step explanation:

Explanation:-

Given data a study of 62 college students finds that their average interest rate is 14 percent with a standard deviation of 9.3 percent.

Sample size 'n' =62

sample mean x⁻ = 14

sample standard deviation 'S' = 9.3

95 percent confidence interval for this single-sample t test

The values are $(x^{-} - t_{0.05} \frac{S}{\sqrt{n} } ,x^{-}+t_{0.05}\frac{S}{\sqrt{n} })$ the 95 percent confidence interval for the population mean 'μ'

Degrees of freedom γ=n-1=62-1=61

t₀.₀₅ = 1.9996 at 61 degrees of freedom

$(14 - 1.9996\frac{9.3}{\sqrt{62} } ,14+1.9996\frac{9.3}{\sqrt{62} })$

(14-2.361 , 14 + 2.361)

[(11.64 , 16.36]

Conclusion:-

95 percent confidence interval for this single-sample t test

[11.64, 16.36]

6 0

2 years ago

ANSWER FOR BRAINLEST (BEST TO EXPLAIN WILL GET IT)

lina2011 [118]

Answer:

-7g+13

Step-by-step explanation:

b) because combining like terms results in:

g-8g=7g

and

15-2= 13

5 0

2 years ago