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2 years ago

Help please What is this expression in simplest form? 4w/w-2 + 3w/w-3

Mathematics

2 answers:

Stels [109]2 years ago

8 0

Answer:

Step-by-step explanation:

7w^2-18w/w^2-5w+6

r-ruslan [8.4K]2 years ago

7 0

Answer:

Step-by-step explanation:

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Solve the equation and enter its solution set below. Enter the solutions in solution set notation separated by a comma and in or

kherson [118]

Answer:

5/6, 7

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3 years ago

Why is it important to know the different properties of three-dimensional figures

IgorLugansk [536]

A two-dimensional shape has length and width. A three-dimensional solid shape also has depth. Three-dimensional shapes, by their nature, have an inside and an outside, separated by a surface. All physical items, things you can touch, are three-dimensional.

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4 years ago

1 1 11 - + - + - 4 3 12 = ????

Temka [501]

Answer:

What?

Step-by-step explanation:

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3 years ago

Read 2 more answers

explain how you could use the greatest common factor of 8 and 20 in the distributive property to write 8 + 20 as a product

GenaCL600 [577]

Answer: 4(5+2) or 4(2+5)

Step-by-step explanation:

8 = 2*2*2

20 = 2*2*5

4(5+2) = 20+8

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3 years ago

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of

freedom are -1.796 & 1.796 with P = 5%}

P(-1.796 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.796) = 0.90

P( $-1.796 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }$ , $21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }$ ]

= [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0

3 years ago