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DanielleElmas [232]
3 years ago
5

PLS HELP TEST ILL GIVE BRAINLIEST

Mathematics
2 answers:
dybincka [34]3 years ago
8 0

Answer:

X must be  negative

because the answer is x=-8

Step-by-step explanation:

Montano1993 [528]3 years ago
3 0
Both answers are negative.
X=-8
X=-2/3
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9.2384E6 in standard form
yan [13]

Answer:3.294. Hope this helps!

Step-by-step explanation:

7 0
3 years ago
Jane buys candy that cost $8 per pound. She will spend more than $48 on candy. What are the possible numbers of pounds she will
Blizzard [7]

Answer:6

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
HELPPPPPP MEEEEE<br><br> y=6x-1<br><br> and -2x-3y=-2 <br><br><br> USING SUBSTITUTION
bazaltina [42]
Answer:
 x = 4 
y =23 

work:
 plug in the value of Y to -2x-3y=-2. -2x - 3(<span>6x-1) = -2
 
multiply 3 and the numbers in parenthesis. -2x -18x - 3 =-2
combine like terms (-2x -18x). -20x - 3 = -2
subtract 3 from both sides to isolate the variable. -20x = -5 
divide to isolate the variable (-20/-5) x = 4
-
now that we know the value of x, it'll be easy to find Y.
Plug it in to the first equation.
</span><span>y=6(4)-1
</span>y = 24 - 1
y = 23

so x = 4 
and y = 23 

hope this helps! :D
7 0
4 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
4 years ago
Energy is required in order to produce any sound.<br><br> A. True<br> B. False
Vikki [24]

Answer: a. True

Step-by-step explanation:

7 0
3 years ago
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