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3 years ago

14

39. If a family borrows $12,381 for an addition to their

1 answer:

Helen [10]3 years ago

4 0

Answer:

206.35

Step-by-step explanation:

5 years = 12x5=60

$12,381 ÷ 60 = $206.35

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(Small sample confidence intervals for a population mean) suppose you are taking a sampling of 15 measurements. you find that x=

Luda [366]

Answer:

The 99% confidence interval is $71.67 < \mu < 78.33$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 15$

The sample mean is $\= x = 75$

The standard deviation is $s = 5$

Given that confidence is 99% then the level of significance is mathematically represented as

$\alpha = 100 - 99$

$\alpha = 1\%$

$\alpha = 0.01$

Next we obtain the critical values of $\frac{ \alpha }{2}$ from the normal distribution table

The value is

$Z_{\frac{ \alpha }{2} } = 2.58$

Generally the margin for error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{ s}{ \sqrt{n} }$

=> $E = 2.58 * \frac{ 5}{ \sqrt{15} }$

=> $E = 3.3307$

The 99% confidence interval is mathematically represented as

$\= x -E < \mu < \= x +E$

=> $75 - 3.3307 < \mu$

=> $71.67 < \mu < 78.33$

3 0

3 years ago

Simplify the expression below and write it as a single logarithm:

OLEGan [10]

The simplification of 3log(x + 4) – 2log(x – 7) + 5log(x - 2) - log(x^2) is $\log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

<u>Solution:</u>

Given, expression is $3 \log (x+4)-2 \log (x-7)+5 \log (x-2)-\log \left(x^{2}\right)$

We have to write in as single logarithm by simplifying it.

Now, take the given expression.

$\rightarrow 3 \log (x+4)-2 \log (x-7)+5 \log (x-2)-\log \left(x^{2}\right)$

Rearranging the terms we get,

$\left.\rightarrow 3 \log (x+4)+5 \log (x-2)-2 \log (x-7)+\log \left(x^{2}\right)\right)$

$\text { since a } \times \log b=\log \left(b^{a}\right)$

$\rightarrow \log (x+4)^{3}+\log (x-2)^{5}-\left(\log (x-7)^{2}+\log \left(x^{2}\right)\right)$

$\text { We know that } \log a \times \log b=\log a b$

$\rightarrow \log \left((x+4)^{3} \times(x-2)^{5}\right)-\left(\log \left((x-7)^{2} \times\left(x^{2}\right)\right)\right.$

$\text { We know that } \log a-\log b=\log \frac{a}{b}$

$\rightarrow \log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

Hence, the simplified form $\rightarrow \log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

4 0

3 years ago

Read 2 more answers

Show that 6^3-1 is divisible by 5 using our identities

Doss [256]

Answer:

Step-by-step explanation:

a³-b³=(a-b)(a²+ab+b²)

6³-1=6³-1³=(6-1)(6²+6*1+1²)=5×43

so 6³-1 is divisible by 5.

8 0

3 years ago

Read 2 more answers

Kevin says these number lines show 7/8 is equivalent to 4/5. Is Kevin correct?

notsponge [240]

Yes, Kevin is correct

3 0

2 years ago

0.6x = 5.4 what is x

zysi [14]

Answer:

x=9

Step-by-step explanation:

5.4 divided by 9 is 0.6

Please mark brainliest

5 0

2 years ago

Read 2 more answers