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3 years ago

10

True or false? The variable in the linear term of a quadratic trinomial is always raised to the first power

1 answer:

Yanka [14]3 years ago

7 0

Answer:

True

Step-by-step explanation:

The <em>first power</em> of the variable is what makes it a <em>linear</em> term.

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Maggie put together 450 party favors for the spring dance. If 170 party favors were donated and the rest were purchased, which e

Troyanec [42]

F

Because p = ammount maggie purchased

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3 years ago

Chatty made this square picture. She wants to glue yarn around the border to make a frame. How much yarn does she need?

Romashka [77]

Answer:

60 inches

Step-by-step explanation:

s = 15 in

Perimeter of a Square = 4 s

= 4 × 15

= 60 in

∴Chatty needs 60 inches of yarn.

Hope that helps!!

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3 years ago

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The cost of an item is 100% of its price what number place of? In the addition problem

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I need the answer too. My brother needs it.

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3 years ago

Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters,

valentinak56 [21]

Answer:

The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the distribution of sample mean is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The information provided is:

<em>μ</em> = 144 mm

<em>σ</em> = 7 mm

<em>n</em> = 50.

Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.

$\bar X\sim N(\mu_{\bar x}=144, \sigma_{\bar x}^{2}=0.98)$

Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:

$P(\bar X-\mu_{\bar x}>2.6)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}} >\frac{2.6}{\sqrt{0.98}})$

$=P(Z>2.63)\\=1-P(Z$

*Use a <em>z</em>-table for the probability.

Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

8 0

3 years ago

Find the dimensions of an open rectangular box with a square base that holds 2000 cubic cm and is constructed with the least bui

Vesnalui [34]

<h3>The dimensions of the given rectangular box are:</h3><h3>L = 15.874 cm , B = 15.874 cm , H = 7.8937 cm</h3>

Step-by-step explanation:

Let us assume that the dimension of the square base = S x S

Let us assume the height of the rectangular base = H

So, the total area of the open rectangular box

= Area of the base + 4 x ( Area of the adjacent faces)

= S x S + 4 ( S x H) = S² + 4 SH ..... (1)

Also, Area of the box = S x S x H = S²H

⇒ S²H = 2000

$\implies H = \frac{2000}{S^2}$

Substituting the value of H in (1), we get:

$A = S^2 + 4 SH = S^2 + 4 S(\frac{2000}{S^2}) = S^2 + (\frac{8000}{S})\\\implies A = S^2 + (\frac{8000}{S})$

Now, to minimize the area put :

$(\frac{dA}{dS} ) = 0 \implies 2S - \frac{8000}{S^2} = 0\\\implies S^3 = 4000\\\implies S = 15.874 \approx 16 cm$

Putting the value of S = 15.874 cm in the value of H , we get:

$\implies H = \frac{2000}{S^2} = \frac{2000}{(15.874)^2} = 7.8937 cm$

Hence, the dimensions of the given rectangular box are:

L = 15.874 cm

B = 15.874 cm

H = 7.8937 cm

3 0

4 years ago