Jan bought 2 novels regularly priced at $24 and $18. Then reduced by 25%. What did she save?

What is the product in simplest form state any restrictions on the variable x^2+9x+18/x+ times x^2-3x-10/x^2+2x-24?

Ostrovityanka [42]

Answer:

X^4+26X^3-24X^2-100 / X^2

(its a fraction btw)

Step-by-step explanation:

ITS THE EQUATION SIMPLIFIED :)

x2+9x+ 18 x x2−3x− 10 x2 +2x−24

= x5+26x4−24x3−10x x3

↓↓↓↓looks like this kinda ↓↓↓↓

x^4+26x^3−24x^2−10

---------------------------------

x^2

6 0

3 years ago

Read 2 more answers

A data set has a median of 12, an upper quartile of 15, a lower quartile of 10, a minimum of 4, and a maximum of 20. Which state

ololo11 [35]

Given:

A data set has a median of 12, an upper quartile of 15, a lower quartile of 10, a minimum of 4, and a maximum of 20.

To find:

The correct statement for the box plot.

Solution:

Lower quartile is 10 and upper quartile is 15, so the box will go from 10 to 15.

Median of the data set is 12, so a line dividing the box will be at 12.

Minimum value is 4 and lower quartile is 10, so the left whisker will go from 4 to 10.

Upper quartile is 15 and maximum value is 20, so the right whisker will go from 15 to 20.

Therefore, the correct option is B.

8 0

3 years ago

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If a builder needs p pieces of wood for every 5s square feet of the building, what is the total amount of lumber of he needs to

Black_prince [1.1K]

Answer: He will need 700 pieces of wood.

Step-by-step explanation: Divide the total by how many square feet one piece of wood will cover, so 3,500/5, and you'll get 700.

Hope this helps.

5 0

3 years ago

What is 7 + 8(2r - 9)

Shalnov [3]

7 + 8(2r - 9)

mutiply the bracket by 8

(8)(2r)= 16r

(8)(-9)= -72

7+16r-72

7-72+16r ( rearranging)

answer:

-65+16r or 16r-65

4 0

3 years ago

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A sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

TEA [102]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = $1,150

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ $1,150

(b) The test statistic is 3.571.

(c) We conclude that the mean of all account balances is significantly different from $1,150.

Step-by-step explanation:

We are given that a sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

We have test the hypothesis to determine whether the mean of all account balances is significantly different from $1,150.

Let $\mu$ = mean of all account balances

(a)So, Null Hypothesis, $H_0$ : $\mu$ = $1,150 {means that the mean of all account balances is equal to $1,150}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ $1,150 {means that the mean of all account balances is significantly different from $1,150}

The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;

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

where, $\bar X$ = sample average balance = $1,200

s = sample standard deviation = $126

n = sample of account balances = 81

(b) So, test statistics = $\frac{1,200-1,150}{\frac{126}{\sqrt{81} } }$ ~ $t_8_0$

= 3.571

The value of the sample test statistics is 3.571.

(c) Now, P-value of the test statistics is given by the following formula;

P-value = P( $t_8_0$ > 3.571) = Less than 0.05%

Because in the t table the highest critical value for t at 80 degree of freedom is given between 3.460 and 3.373 at 0.05% level.

Now, since P-value is less than the level of significance as 5% > 0.05%, so we sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean of all account balances is significantly different from $1,150.

8 0

3 years ago