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evablogger [386]
3 years ago
14

Question 4 (1 point)

Mathematics
1 answer:
kow [346]3 years ago
8 0

Answer:

pretty sure its A n B = {1,5}

Step-by-step explanation:

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95435.20 miles

Step-by-step explanation:

just add them both together

7 0
2 years ago
Lori wants to trade more than 3/10 but less than 4/5 of her stickers.
mote1985 [20]

Answer:

7/10, 6/10, or 5/10

Step-by-step explanation:

So, first, we change 4/5 into tenths. That equals 8/10. 3/10 and 8/10.

6 0
2 years ago
Find an equivalent ratio in simplest terms: 56:18
eimsori [14]

Answer: 28:9

Step-by-step explanation:

56:18

56÷2=28

18÷2=9

=28:9

8 0
3 years ago
Read 2 more answers
Write the equation of each graph after the indicated transformations. The graph of y = √x is reflected in the x-axis, stretched
Zarrin [17]

Answer:

can you mark me of a brainliest please answers to you is C

7 0
2 years ago
A sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.
TEA [102]

Answer:

(a) Null Hypothesis, H_0 : \mu = $1,150  

    Alternate Hypothesis, H_A : \mu \neq $1,150

(b) The test statistic is 3.571.

(c) We conclude that the mean of all account balances is significantly different from $1,150.

Step-by-step explanation:

We are given that a sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

We have test the hypothesis to determine whether the mean of all account balances is significantly different from $1,150.

<u><em>Let </em></u>\mu<u><em> = mean of all account balances</em></u>

(a)So, Null Hypothesis, H_0 : \mu = $1,150     {means that the mean of all account balances is equal to $1,150}

Alternate Hypothesis, H_A : \mu \neq $1,150    {means that the mean of all account balances is significantly different from $1,150}

The test statistics that will be used here is <u>One-sample t test statistics</u> as we don't know about population standard deviation;

                               T.S.  = \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample average balance = $1,200

             s = sample standard deviation = $126

             n = sample of account balances = 81

(b) So, <u>test statistics</u>  =  \frac{1,200-1,150}{\frac{126}{\sqrt{81} } }  ~ t_8_0

                               =  3.571

The value of the sample test statistics is 3.571.

(c) <u>Now, P-value of the test statistics is given by the following formula;</u>

          P-value = P( t_8_0 > 3.571) = <u>Less than 0.05%</u>

Because in the t table the highest critical value for t at 80 degree of freedom is given between 3.460 and 3.373 at 0.05% level.

Now, since P-value is less than the level of significance as 5% > 0.05%, so we sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean of all account balances is significantly different from $1,150.

8 0
3 years ago
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