May anyone help me on this one

Simplify the expression.

Mashcka [7]

Answer:

-2.5

Step-by-step explanation:

-0.25x-6-5.5x-8.4=0

-5.75x-14.4=0

-5.75x=14.4

x= -2.5

3 0

3 years ago

Whats the answer??? Pleaseeee!!!!

kvasek [131]

To find the 20th term in this sequence, we can simply keep on adding the common difference all the way until we get up to the 20th term.

The common difference is the number that we are adding or subtracting to reach the next term in the sequence.

Notice that the difference between 15 and 12 is 3.

In other words, 12 + 3 = 15.

That 3 that we are adding is our common difference.

So we know that our first term is 12.

Now we can continue the sequence.

12 ⇒ 1st term

15 ⇒ 2nd term

18 ⇒ 3rd term

21 ⇒ 4th term

24 ⇒ 5th term

27 ⇒ 6th term

30 ⇒ 7th term

33 ⇒ 8th term

36 ⇒ 9th term

39 ⇒ 10th term

42 ⇒ 11th term

45 ⇒ 12th term

48 ⇒ 13th term

51 ⇒ 14th term

54 ⇒ 15th term

57 ⇒ 16th term

60 ⇒ 17th term

63 ⇒ 18th term

66 ⇒ 19th term

69 ⇒ 20th term

This means that the 20th term of this arithemtic sequence is 69.

5 0

3 years ago

Anne is checking out of a hotel. The price

kolezko [41]

The tax is 2.88 so it would be 48.00 +2.88=50.88
She will pay $50.88 with tax

3 0

2 years ago

NEED HELP Fast open up the Screenshot

Evgesh-ka [11]

I think the answer is A but I am not sure. I think you would take the opposite of the y axis.

8 0

3 years ago

Read 2 more answers

A average amount of money on deposit in a savings account is $7500. Suppose a random sample of 49 accounts shows the average in

Valentin [98]

Answer:

z=-3.79

$p_v =2*P(z$

Step-by-step explanation:

1) Data given and notation

$\bar X=6850$ represent the mean average amount of money on deposit in savings account for the sample

$\sigma=1200$ represent the population standard deviation for the sample

$n=49$ sample size

$\mu_o =7500$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the amount of money on deposit in savings account differs from 7500, the system of hypothesis would be:

Null hypothesis: $\mu =7500$

Alternative hypothesis: $\mu \neq 7500$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{6850-7500}{\frac{1200}{\sqrt{49}}}=-3.79$

4)P-value

Since is a two-sided test the p value would be:

$p_v =2*P(z$

5) Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average amount of money deposit in savings account its significantly different from 7500 tons at 1% of signficance.

7 0

3 years ago