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2 years ago

Two other ways to write 0.632

2 answers:

6 0

*expanded form: 6/10 + 3/100 + 2/1000 (that is not division those r fractions!)
*written: six hundred thirty-two thousandTHS

always put "ths" when it is after a decimal point.

NeX [460]2 years ago

3 0

As a fraction 632/1,000 and as a percentage 63.2%

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lukranit [14]

Answer:

500 together and they are all the same

Step-by-step explanation:

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2 years ago

Can someone help me please

lukranit [14]

Answer:

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Step-by-step explanation:

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2 years ago

Can you simplify the equation 64x(squared)-112x+49

Leni [432]

Answer:

You cannot simply 64x^2-112x+ 49 further than it is already simplified.

Step-by-step explanation:

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2 years ago

Given the figure below, calculate the area

Aleksandr [31]

Answer:

1+√3

Step-by-step explanation:

Area of the triangle = 1/2 * base * height

Area of the triangle = 1/2 * AC * BD

We need to get AC

AC = AD+DC

Get base AD using SOH CAH TOA

tan 45 = 4/AD

AD = 4/tan45

AD = 4/1

AD = 4

Get DC

tan 30 = 4/DC

DC = 4/tan 30

DC = 4/(1/√3)

DC = 4√3

AC = 4 + 4√3

AC = 4(1+√3)

Get the Area of the triangle

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Area of the triangle = 1+√3

6 0

2 years ago

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:

The question is incomplete:

"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."

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