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2 years ago

I need the answer to this, i don’t understand this .

Mathematics

1 answer:

tekilochka [14]2 years ago

7 0

The answer is c........

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jacks car used 18 gallons of gasoline to go 468 miles. At that rate, how many gallons would be used to go 754 miles? Explain you

matrenka [14]

Well if you take 468 and divide it by 18 you get 26 that 26mpg take 754 divide it by 26 and you get 29 and that's how many gallons he used

6 0

2 years ago

Read 2 more answers

If the perimeter of an equilateral triangle is 30cm, find its area.

lora16 [44]

Answer:

A ≈ 43.3 cm²

Step-by-step explanation:

the area (A) of an equilateral triangle is calculated as

A = $\frac{s^2\sqrt{3} }{4}$ ( s is a side of the triangle )

given perimeter = 30 cm , then

s = 30 cm ÷ 3 = 10 cm

then

A = $\frac{10^2\sqrt{3} }{4}$ = $\frac{100\sqrt{3} }{4}$ = 25 $\sqrt{3}$ ≈ 43.3 cm² ( to the nearest tenth )

7 0

2 years ago

Need help anything helps thanks

yKpoI14uk [10]

A binomial squared is written like

$x^2+2ax+a^2$

So, we have $2a=6 \iff a=3$

So, we want to complete the square $(x+3)^2 = x^2+6x+9$

We can add and subtract 9 to write

$x^2+6x = x^2+6x+(9-9) = (x^2+6x+9)-9 = (x+3)^2-9$

5 0

2 years ago

Which fraction is equivalent

Alina [70]

The answer would be d

3 0

1 year ago

Read 2 more answers

Let us suppose we have data on the absorbency of paper towels that were produced by two different manufacturing processes. From

maksim [4K]

Answer:

The 95% CI for the difference of means is:

$-155.45 \leq \mu_1-\mu_2 \leq -44.55$

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>"Find a 95% confidence interval on the difference of the towels mean absorbency produced by the two processes. Assumed that the standard deviations are estimated from the data. Round to two decimals places."</em>

Process 1:

- Sample size: 10

- Mean: 200

- S.D.: 15

Process 2:

- Sample size: 4

- Mean: 300

- S.D.: 50

The difference of the sample means is:

$M_d=M_1-M_2=200-300=-100$

The standard deviation can be estimated as:

$\sigma_d=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\\\\sigma_d=\sqrt{\frac{15^2}{10}+\frac{50^2}{4}} =\sqrt{22.5+625}=\sqrt{647.5}=25.45$

The degrees of freedom are:

$df=n_1+n_2-2=10+4-2=12$

The t-value for a 95% confidence interval and 12 degrees of freedom is t=±2.179.

Then, the confidence interval can be written as:

$M_d-t\cdot \sigma_d\leq \mu_1-\mu_2 \leq M_d+t\cdot \sigma_d\\\\-100-2.179*25.45\leq \mu_1-\mu_2 \leq -100+2.179*25.45\\\\-100-55.45 \leq \mu_1-\mu_2 \leq -100+55.45\\\\ -155.45 \leq \mu_1-\mu_2 \leq -44.55$

8 0

2 years ago