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jasenka [17]
3 years ago
14

What number does a Roman numerals MDC XL IX represent​

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
7 0

Answer:

the answer is 1649

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PLEASE HELPPP!! A box contains only black and white chips There are 8 black chips and 6 white chips How many white chips must be
Law Incorporation [45]
Lets take away 2 white chips...

8 black and 4 white...total of 12 chips
P(black) = 8/12 = 2/3

remove 2 white chips <==
3 0
3 years ago
Please help ill give brainliest pls dont guess
LiRa [457]
When y=2 and y=5

1. 2y-1 and (3y-5+y or 4y-5)
when y=2 ; 2(2)-1 = 3 and 4(2)-5=3
when y=5 ; 2(5)-1 = 9 and 4(5)-5=15

----nonequivalent-----

2.5y+4 and (7y+4-2y or 5y+4)
so you don't have to place any value in because 5y+4 and 7y+4-2y are equal,
whatever you place any value in, it will be all the same then

-----equivalent------

and no need to find more
7 0
3 years ago
PLEASE HELP!!!!!!! A family has two cars. The first car has a fuel efficiency of 35 miles per gallon of gas and the second has a
Marianna [84]

Answer:

The answer to your question is car 1 = 30 gal and car 1 = 20 gal

Step-by-step explanation:

car 1 = a

car 2 = b

Efficiency of car 1 = 35 mi/gal

Efficiency for car 2 = 20 mi/gal

Total distance = 1450

Total gas consumption = 50 gal

Equations

                          35a + 20b = 1450            ------- (I)

                             a   +     b =  50               ------- (II)

Solve by elimination

Multiply equation II by -35

                          35a  + 20b  = 1450

                         -35a  - 35b   = -1750

Simplify

                            0    - 15b    =  -300

Solve for b

                                        b =  -300/-15

Result

                                        b = 20

Substitute b in equation II to find a

                               a +  20  = 50

Solve for a

                               a  = 50 -20

Result

                               a  = 30

4 0
3 years ago
What is the constant of vanation,<br>of the line y = for through (3.18) and (5,30)?​
Naily [24]

the constant of variation or namely its slope will be

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{18})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{30}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{30-18}{5-3}\implies \cfrac{12}{2}\implies 6

8 0
3 years ago
Members of the millennial generation are continuing to be dependent on their parents (either living with or otherwise receiving
Morgarella [4.7K]

Answer:

a)

\bf H_0: The mean of adults aged 18 to 32 that continue to be  dependent on their parents is 0.3

\bf H_a: The mean of adults aged 18 to 32 that continue to be  dependent on their parents is greater than 0.3

b) 34%

c) practically 0

d) Reject the null hypothesis.

Step-by-step explanation:

a)

Since an individual aged 18 to 32 either continues to be dependent on their parents or not, this situation follows a Binomial Distribution and, according to the previous research, the probability p of “success” (depend on their parents) is 0.3 (30%) and the probability of failure q = 0.7

According to the sample, p seems to be 0.34 and q=0.66

To see if we can approximate this distribution with a Normal one, we must check that is not too skewed; this can be done by checking that np ≥ 5 and nq ≥ 5, where n is the sample size (400), which is evident.

<em>We can then, approximate our Binomial with a Normal </em>with mean

\bf np = 400*0.34 = 136

and standard deviation

\bf \sqrt{npq}=\sqrt{400*0.34*0.66}=9.4742

Since in the current research 136 out of 400 individuals (34%) showed to be continuing dependent on their parents:

\bf H_0: The mean of adults aged 18 to 32 that continue to be  dependent on their parents is 0.3

\bf H_a: The mean of adults aged 18 to 32 that continue to be  dependent on their parents is greater than 0.3

So, this is a r<em>ight-tailed hypothesis testing. </em>

b)

According to the sample the proportion of "millennials" that are continuing to be dependent on their parents is 0.34 or 34%

c)

Our level of significance is 0.05, so we are looking for a value \bf Z^* such that the area under the Normal curve to the right of \bf Z^* is ≤ 0.05

This value can be found by using a table or the computer and is \bf Z^*= 1.645

<em>Applying the continuity correction factor (this should be done because we are approximating a discrete distribution (Binomial) with a continuous one (Normal)), we simply add 0.5 to this value and </em>

\bf Z^* corrected is 2.145

Now we compute the z-score corresponding to the sample

\bf z=\frac{\bar x -\mu}{s/\sqrt{n}}

where  

\bf \bar x= mean of the sample

\bf \mu= mean of the null hypothesis

s = standard deviation of the sample

n = size of the sample

The sample z-score is then  

\bf z=\frac{136 - 120}{9.4742/20}=16/0.47341=33.7759

The p-value provided by the sample data would be the area under the Normal curve to the left of 33.7759 which can be considered zero.

d)

Since the z-score provided by the sample falls far to the left of  \bf Z^* we should reject the null hypothesis and propose a new mean of 34%.

7 0
3 years ago
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