Expand and simplify 5(2x+1)-2(3x-6)

What expressions are equal to 16.04

quester [9]

where are the expressions?

7 0

3 years ago

Drag each tile to the correct box.<br>Order the expressions from least value to greatest value.<br>

STatiana [176]

Answer:

-2 19/20, 2 3/5. 3.125, 10.35

Step-by-step explanation:

Let's solve each expression:

Solve:

-4.5 x -2.3

Multiply:

-4.5 x -2.3

10.35

Therefore, the simplified expression is 10.35

Solve:

5/-1.6

Divide:

5/-1.6

3.125

Therefore, the simplified expression is 3.125

Solve:

5 6/15 + (-2 4/5)

First, make the denominators of 4/5 equal to 15:

4/5 = ?/15

Multiply both the numerator and denominator by 3:

4 x 3 = 12

5 x 3 = 15

The new fraction is 12/15.

Add:

5 6/15 + (-2 12/15)

2 9/15

Simplify:

2 9/15 = 2 3/5

(divide the numerator and denominator by 3)

Therefore, the simplified expression is 2 and three-fifths.

Solve:

-3 3/10 - (-7/20)

First, make the denominators of 3/10 equal to 20:

3/10 = ?/20

Multiply both the numerator and denominator by 2:

3 x 2 = 6

10 x 3 = 20

The new fraction is 6/20.

Subtract:

-3 6/20 - (-7/20)

-2 19/20

Therefore, the simplified expression is negative 2 and nineteen-twentieths.

Order from least to greatest:

-2 19/20, 2 3/5. 3.125, 10.35

3 0

3 years ago

An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by th

monitta

Answer:

a) n =1623

b) $ME=2.0538\sqrt{\frac{0.21 (1-0.21)}{1623}}=0.0208$

Step-by-step explanation:

1) Notation and definitions

$n$ random sample taken

$\hat p=0.19$ estimated proportion of customers are not satisfied with the service provided by the local dealer

Confidence =0.96 or 96%

Me= 0.02 represent the margin of error

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 96% of confidence, our significance level would be given by $\alpha=1-0.96=0.04$ and $\alpha/2 =0.02$ . And the critical value would be given by:

$z_{\alpha/2}=-2.0538, z_{1-\alpha/2}=2.0538$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.02$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.19(1-0.19)}{(\frac{0.02}{2.0538})^2}=1622.912$

And rounded up we have that n=1623

Part b

The margin of error is given by:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

So then if w replace the value of n obtained from part a we got:

$ME=2.0538\sqrt{\frac{0.21 (1-0.21)}{1623}}=0.0208$

3 0

3 years ago

The diner sells bagels for $0.75 each. the bagel shop sells 1 dozen for $7.00. which is the better pric

Varvara68 [4.7K]

The bagel shop has the better price because 1 dozen is 12 bagels and if you multiply 0.75 x 12 = 9.00
so the better price is 7.00

6 0

3 years ago

PLEASE HELP ME WITH THIS I WILL GIVE BRAINLIEST.

Elena-2011 [213]

Answer:

1007

Step-by-step explanation:

The volume of the cylindrical storage tank is : π·40(squared)·60 = approximately 302000.

The volume of a rectangular prism is : 5 x 6 x 10 = 300

To catch all the leaking liquids, you need : 302000 : 300 = 1006.6666667 (round to 1007) rectangular prisms

4 0

2 years ago

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