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kodGreya [7K]
3 years ago
7

If a student did not travel to a Spanish-speaking country, how many times more likely is it that the student did not take Spanis

h?
It is 22 times as likely.
It is 9 times as likely.
It is 4.5 times as likely.
It is 2.2 times as likely.
Mathematics
1 answer:
olasank [31]3 years ago
3 0

Answer:

It is 2.2 times as likely.

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Find the value of x and y<br> 3х<br> 20<br> 25<br> 18<br> бу -2<br> Х=<br> ү=
sweet-ann [11.9K]

Answer: first one =109360044xy−36453348x−2, 2nd,=3037778998,last one,=1968480790

Step-by-step explanation:Evaluate for x=3x,y=6y−2

33x(2025186)(6y−2)−2

33x(2025186)(6y−2)−2

=109360044xy−36453348x−2

Evaluate for x=20,y=25

(3)(20)(2025186)(25)−2

(3)(20)(2025186)(25)−2

=3037778998

because 18 is by itself i just did Evaluate for x=18,y=18

(3)(18)(2025186)(18)−2

(3)(18)(2025186)(18)−2

=1968480790

7 0
2 years ago
Factor the common factor out of the expression 8+20a+18a^2
ruslelena [56]
First let's rewrite it...

18a^2 + 20a + 8

Factor out a 2

2 (9a^2 + 10a + 4)
5 0
2 years ago
A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past
Vadim26 [7]

Answer:

We conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

Step-by-step explanation:

We are given that a particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8000.

From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9800 miles.

<u><em>Let </em></u>\mu<u><em> = average miles for deluxe tires</em></u>

So, Null Hypothesis, H_0 : \mu \geq 50,000 miles   {means that deluxe tire averages at least 50,000 miles before it needs to be replaced}

Alternate Hypothesis, H_A : \mu < 50,000 miles    {means that deluxe tire averages less than 50,000 miles before it needs to be replaced}

The test statistics that will be used here is <u>One-sample z test statistics</u> as we know about population standard deviation;

                                  T.S.  = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean lifespan = 46,500 miles

            \sigma = population standard deviation = 8000 miles

            n = sample of tires = 28

So, <u><em>test statistics</em></u>  =  \frac{46,500-50,000}{\frac{8000}{\sqrt{28} } }

                               =  -2.315

The value of the test statistics is -2.315.

Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z as -2.315 < -1.6449, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

4 0
2 years ago
Match the multiplication Expressoins to the exponential expressions ​
Harman [31]

Answer:

Step-by-step explanation:

The pattern is easy to see.

8 0
2 years ago
Answer C and D please
DENIUS [597]

Answer:

C. solution:

3(3c-2) = 5(2c-1)

or, 9c-6 = 10c-5

or, -6+5 =<em> </em>10c-9c

or, -1 = 1c

Hence, c = -1.

D. solution:

5(5-2a) = 4(6-a)

or, 25-10a = 24 - 4a

or, 25-24 = -4a+10a

or, 1 = 6a

or, 1/6 = a

Hence, a = 1/6.

6 0
2 years ago
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