All categories

3 years ago

5X23X2 using properties to find the sum or product

Mathematics

1 answer:

lianna [129]3 years ago

4 0

$5\cdot 23 \cdot 2=(5\cdot 2)\cdot 23 = 10\cdot 23=230$

You might be interested in

(x^5 y)/(x^2 y^6) please help i dont understand this

Dvinal [7]

Answer:

x³/y⁵

Simplify the expression.

5 0

3 years ago

Which expression is equivalent to 8x - 12y +32

prisoha [69]

The answer is shown above

6 0

2 years ago

Read 2 more answers

Help me please, I’m stuck!

Veseljchak [2.6K]

Answer:

5. (x, y) ⇒ (x +5, y +1)

6. (x, y) ⇒ (x +4, y -1)

Step-by-step explanation:

A translation rule will generally have the form ...

(x, y) ⇒ (x +h, y +k)

where (h, k) is the horizontal and vertical distance to the right and up that the figure is being translated.d

5) The image is 5 units right and 1 unit up from the original, so in the above formula, (h, k) = (5, 1) and the translation rule is ...

(x, y) ⇒ (x +5, y +1)

6) The image is 4 units right and 1 unit down from the original, so (h, k) = (4, -1) and the transformation rule is ...

(x, y) ⇒ (x +4, y -1)

4 0

3 years ago

Solving Exponential and Logarithmic Equations In Exercise, solve for x.<br> 500(1.075)120x = 100.000

Volgvan

Answer:

The solution is:

$x = 0.61$

Step-by-step explanation:

The first step to solve this equation is placing everything with the exponential to one side of the equality, and everything without the exponential to the other side. So

$500(1.075)^{120x} = 100000$

$(1.075)^{120x} = \frac{100000}{500}$

$(1.075)^{120x} = 200$

To find x, we have to apply log to both sides of the equality.

We also have that:

$\log{a^{x}} = x\log{a}$

$\log{(1.075)^{120x}} = \log{200}$

$120x\log{1.075} = 2.30$

$120x*0.03 = 2.30$

$3.77x = 2.30$

$x = \frac{2.30}{3.77}$

$x = 0.61$

4 0

3 years ago

Construct a 95% confidence interval for the population mean, μ. Assume the population has a normal distribution. A random sample

GuDViN [60]

Answer: $(628.48,\ 661.52)$

Step-by-step explanation:

Given : Sample size : $n=16$ , which is a small sample (, 30) so we use t-test.

Sample mean : $\overline{x}=645 \text{ hours}$

Standard deviation : $\sigma = 31 \text{ hours}$

Significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=2.131$

The confidence interval for population mean is given by :-

$\overline{x}\pm t_{n-1,\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=645\pm(2.131)\dfrac{31}{\sqrt{16}}\\\\\approx645\pm16.52\\\\=(645-16.52,\ 645+16.52)\\\\=(628.48,\ 661.52)$

Hence, a 95% confidence interval for the population mean $\mu$ = $(628.48,\ 661.52)$

6 0

3 years ago