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pshichka [43]
3 years ago
12

Employees at a company are given £1,200 to spend on items for the office.

Mathematics
1 answer:
Vesna [10]3 years ago
7 0

Answer:

100 $

Step-by-step explanation:

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saul85 [17]

Answer: 120

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=

1.2

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6 0
3 years ago
A phone company charges 0.1 per minutes talked and additional $24 for services each month. Write the liner function C(n) where n
ella [17]

Answer:

the cost of talking for 15 minutes would be $1.5

15x0.1=1.5

for $50 you could talk for 500 minutes.

50/0.1=500

double check, 500x0.1=50

6 0
3 years ago
Which shows a rational number plotted correctly on a number line?
irinina [24]

Answer:

Top option

Step-by-step explanation:

-19\frac{1}{2} = -\frac{39}{2}

According to this, this is definitely in the right place.

I am joyous to assist you anytime.

7 0
3 years ago
Read 2 more answers
What is the value of x in the inequality
GalinKa [24]

<u>Answer & Step-by-step explanation:</u>

<u>Question 18 </u>

Isolate the variable by dividing each side by factors that don't contain the variable.

Inequality Form: x ≤ −10

Interval Notation: (−∞ , −10]

<u>Question 19</u>

Isolate the variable by dividing each side by factors that don't contain the variable.

Inequality Form: x ≤ 4

Interval Notation: (−∞ , 4]

8 0
3 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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