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2 years ago

10

1. 5 less than one fourth of x is y.

1 answer:

horsena [70]2 years ago

6 0

Answer

y = (1/4)x - 5 or 4y - x + 20 = 0

Explanation

To get the rule required, we interpret the sentence and write it in a mathematical form.

5 less than one fourth of x is y

one fourth of x ⇒ (1/4)x

5 less than one fourth of x is y ⇒ (1/4)x - 5 = y

y = (1/4)x - 5

4y = x - 20

4y - x + 20 = 0

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Answer:

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3 years ago

What is FOIL and PEMDAS?

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FOIL and PEMDAS are both ways to solve mathematical equations.

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3 0

2 years ago

Read 2 more answers

Simplify ten to the eighth divided by ten to the negative third.

In-s [12.5K]

Answer:

$10^{11}$

Step-by-step explanation:

A statement is given i.e. "ten to the eighth divided by ten to the negative third."

It means that 10 to power 8 divided by 10 to the power -3.

Mathematically,

$\dfrac{10^8}{10^{-3}}$

We know that, $\dfrac{x^a}{x^b}=x^{a-b}$

Here, x = 10, a = 8 and b = -3

So,

$\dfrac{10^8}{10^{-3}}=10^{8-(-3)}\\\\=10^{8+3}\\\\=10^{11}$

So, the answer is $10^{11}$

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3 years ago

Clayton is solving the equation 5x^2 = 20 – 12x with the quadratic formula.

Stolb23 [73]

Answer:

$a=5, b=20, c= -12$

Step-by-step explanation:

4 0

2 years ago

The average hourly wage for a high school student in Illinois is $8.25 per hour with a standard deviation of $.25. Assume hourly

serious [3.7K]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is

X: hourly wage of a high school student

X~N(μ;σ²)

X[bar]= $8.25

σ= $0.25

For a 95% CI for the average hourly wage for<u> 1 </u>high school student:

For this exercise you have to consider the sample size n=1

a) The margin of error of the CI is calculated as:

d= $Z_{1-\alpha /2}$ * σ

1-α: 0.95 ⇒ α:0.05

1-α/2:0.975

$Z_{1-\alpha /2}= Z_{0.975}= 1.96$

d= 1.96 * 0.25= 0.49

b)

The formula for the interval is

X[bar] ± $Z_{1-\alpha /2}$ * σ

[8.25 ± 0.49]

[7.76; 8.74]

With a 95% you'd expect the interval $[7.76; 8.74] to include the average hourly wage of one high school student in Illinois.

c)

Te calculated interval is $ [7.76; 8.74] the value $9.15 does nor fall within that acceptable range.

I hope this helps!

7 0

3 years ago