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2 years ago

Please show work thank you math experts

Mathematics

1 answer:

MakcuM [25]2 years ago

3 0

Answer:

See below

Step-by-step explanation:

1) The distribution is skewed left because the majority of the data lies in the left tail.

2) Given what we answered in #1, the mean is going to be the closest of the three measures of central tendency, with the median behind the mean, and the mode behind the median.

3) An outlier is a piece of data that is furthest away from of a set of observed data.

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Can u solve this for me

Studentka2010 [4]

X^2 + 4x + 16 is the answer

8 0

3 years ago

Lake has a total of $10,000 to invest in two accounts. One account earns 1% simple interest, and the other earns 10% simple inte

Elza [17]

Answer:

$8000 at 1%
$2000 at 10%

Step-by-step explanation:

It often works well to let a variable represent the amount invested at the higher rate. Then an equation can be written relating amounts invested to the total interest earned.

<h3>setup</h3>

Let x represent the amount invested at 10%. Then 10000-x is the amount invested at 1%. The total interest earned is ...

0.10x +0.01(10000 -x) = 280

<h3>solution</h3>

Simplifying gives ...

0.09x +100 = 280

0.09x = 180 . . . . . . . subtract 100

x = 2000 . . . . . . divide by 0.09

10000 -x = 8000 . . . . amount invested at 1%

$8000 should be invested in the 1% account

$2000 should be invested in the 10% account

5 0

2 years ago

The length and width of a rectangle are measured as 55 cm and 49 cm, respectively, with an error in measurement of at most 0.1 c

valentina_108 [34]

Answer:

The maximum error in the calculated area of the rectangle is $10.4 \:cm^2$

Step-by-step explanation:

The area of a rectangle with length $L$ and width $W$ is $A= L\cdot W$ so the differential of <em>A</em> is

$dA=\frac{\partial A}{\partial L} \Delta L+\frac{\partial A}{\partial W} \Delta W$

$\frac{\partial A}{\partial L} = W\\\frac{\partial A}{\partial W}=L$ so

$dA=W\Delta L+L \Delta W$

We know that each error is at most 0.1 cm, we have $|\Delta L|\leq 0.1$ , $|\Delta W|\leq 0.1$ . To find the maximum error in the calculated area of the rectangle we take $\Delta L = 0.1$ , $\Delta W = 0.1$ and $L=55$ , $W=49$ . This gives