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3 years ago

Make 54/2 to a mixed numbers

Mathematics

1 answer:

Burka [1]3 years ago

4 0

Answer: 27

Step-by-step explanation: Look below for an attached explanation.

Hope this helps!

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What is 83 x 96 fzoigjwcsvgfmcgksjkvzmclkm

nasty-shy [4]

7968 because you multiply the two numbers to get the final answer

4 0

3 years ago

Justyna invested $1500 in an account at a 2.5% interest rate compounded annually. She made no deposits or withdrawals on the

olga55 [171]

Answer:

$1783.03

Step-by-step explanation:

Annually compounding interest formula:

PV(1+i)ⁿ

1500(1+.025)⁷

1500(1.025)⁷

1783.028631

which rounds to

1783.03

4 0

2 years ago

Read 2 more answers

The circumference of a circle is 6 pi cm what is the length of an arc of 140 degree

ValentinkaMS [17]

Since the total circumference of the circle is 60 pi cm which passes around 360˚, the length of an arc of 140˚ will be equal to 140/360 = 7/18 times the total length:
l = (7/18) * 60 pi cm = 23.3 pi cm

3 0

3 years ago

Need help asap will give brain and all non helpful answers will be reported

jonny [76]

Answer:

Local max is .5

Step-by-step explanation:

The local maximum is for x values between 2 and 4

It looks like the x value is 2.5 and the y value is .5

5 0

3 years ago

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .

$39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

$-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
$-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

$(-20.5707,-11.4293)$

If we want to interpret the confidence interval of (-20.5707, -11.4293), we can say...

We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.

5 0

1 year ago