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3 years ago

Find The Measure Of "X"

Mathematics

1 answer:

murzikaleks [220]3 years ago

7 0

Answer:

mesure of x find it hahhahaha joke lng mga lodss

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Write the equation of the circle graphed below. Center = (1,-1) radius = 0.5

spin [16.1K]

Answer:

Step-by-step explanation:

u h h a y

h i o r o

w e u

4 0

3 years ago

Which statement best describes the use of a simulation to predict the probability that two randomly chosen people will both have

laila [671]

<span>Randomly generate an integer from 1 to 7 two times, and the probability is 1/7 ^2

This is the </span><span>statement that best describes the use of a simulation to predict the probability that two randomly chosen people will both have their birthdays on a Monday.

There are 7 days in a week, so there are 7 choices but only 1 Monday. So, 1/7 is the probability that a person's birthday falls on a Monday.

1st person asked will have 1/7 probability.
2nd person asked will also have 1/7 probability

So, (1/7)</span>² is the probability that both persons will have their birthdays on a Monday.

8 0

3 years ago

Read 2 more answers

Solve for x. Show your Work.

andreyandreev [35.5K]

Answer:

50+3x+4 = 90

54+3x = 90

3x = 90-54

3x = 36

x = 12

3x+4 = 3(12)+4

= 36+4

= 40

5 0

3 years ago

The bad debt ratio for a financial institution is defined to the dollar value of loans defaulted divided by the total dollar val

vlabodo [156]

Answer:

Step-by-step explanation:

From the information given:

Consider X to be the random variable denoting the bad debt ratios for Ohio Bank.

Then, $X \sim N ( \mu, \sigma ^2)$

Thus the null hypothesis and the alternative can be computed as:

Null hypothesis:

$H_o : \mu \leq3.5\%$

Alternative hypothesis

$H_1 : \mu > 3.5\%$

The type I and type II error is as follows:

Type I:

The mean bad debt ratio is > 3.5% when it is not

Type II:

The mean bad debt ratio is ≤ 3.5% when it is not.

The test statistics can be calculated by using the formula:

$t = \dfrac{\overline x - \mu}{\dfrac{\sigma}{\sqrt{n}}}$

where;

sample size n = 7

mean = 6+8+5+9+7+5+8 = 48

sample mean $\overline x =\dfrac{48}{7}$

$\overline x$ = 6.86

sample standard deviation is :

$s = \sqrt{\dfrac{\sum( x -\overline x)^2}{n-1}}$

$s = \sqrt{\dfrac{( 6 -6.86)^2+( 8-6.86)^2+ ( 5 -6.86)^2 + ...+( 7 -6.86)^2+ ( 5 -6.86)^2+( 8 -6.86)^2 }{7-1}}$

s = 1.573

population mean $\mu = 3.5$

Therefore, the test statistics is :

$t = \dfrac{\overline x - \mu}{\dfrac{\sigma}{\sqrt{n}}}$

$t = \dfrac{6.86- 3.5}{\dfrac{1.573}{\sqrt{7}}}$

$t = \dfrac{3.36}{\dfrac{1.573}{2.65}}$

$t = \dfrac{3.36\times {2.65}}{{1.573}}$

t = 5.660

At significance level of 0.01

$t_{0.01} = 3.707$

P - value = P(T > 5.66)

P - value = 1 - (T < 5.66)

P - value = 1 - 0.9993

P-value = 0.0007

Therefore, since $t_{0.01} < t$ , we reject the null hypothesis and conclude that the claim that the mean bad debt ratio for Ohio banks is higher than the mean for all financial institutions is true.

4 0

4 years ago

The express the expression 6×10 to the -7÷3×10 to the -3rd is equivalent to what?

vlabodo [156]

$(6\times10^{-7})\div(3\times10^{-3})=\dfrac{6\times10^{-7}}{3\times10^{-3}}=\dfrac{6}{3}\times10^{-7-(-3)}=2\times10^{-7+3}\\\\=\boxed{2\times10^{-4}}\\\\Used:\\\\\dfrac{a^n}{a^m}=a^{n-m}$

7 0

3 years ago