All categories

3 years ago

In the arithmetic sequence, -12, -17, -22, -27... which term has a value of -68?

Mathematics

1 answer:

RSB [31]3 years ago

7 0

Answer:

-68 is not a term in the given sequence

Step-by-step explanation:

The given sequence is -12, -17, -22, -27...

There is a common difference of d=-17--12=-5

The explicit rule for this sequence is $f(n)=-12-5(n-1)$

Or $f(n)=-5n-7$

To find the term that has a value of -68 we equate the explicit formula and solve for n.

$-68=-5n-7$

$\implies -68+7=-5n$

$\implies -61=-5n$

$n=\frac{61}{5} =12.2$

The position should be a natural number.

Since n is a decimal, it tells us that -68 is not a term in the given sequence

You might be interested in

Equation of the line passing through (7,5) and (1, 2)

Trava [24]

Answer:

$y = \frac{1}{2}x + \frac{3}{2}$

Step-by-step explanation:

Assuming this is a linear equation in the form of $y = mx + b$ , we can first solve for the slope using $\frac{y_2 - y_1}{x_2 - x_1}$ , for these points, the slope would be $\frac{1}{2}$ . Now we plug in the points to get the equation:

plugging in (7,5), we get $5 = \frac{1}{2}$ × $7$ + $b$

and solving for b, we get b = $\frac{3}{2}$

therefore, the equation of the line passing through (7,5) and (1,2) is $y = \frac{1}{2}x + \frac{3}{2}$ .

hope this helps!

6 0

2 years ago

Solve the equation. Check your answer. 6x - 3= 5x +5 X=

ZanzabumX [31]

Answer:

x = 8

Step-by-step explanation:

6x - 3 = 5x + 5

Move variable to the left hand side and change their sign.

$\sf \: 6x - 5x - 3 = 5$

Calculate like terms.

x - 3 = 5.

Move constant to the right hand side and change their sign.

x = 5 + 3
x = 8

Check our answer :-

6x -3 = 5x + 5

plug the 8 as x.

6 ( 8 ) - 3 = 5 ( 8 ) + 5
48 - 3 = 40 + 5
45 = 45

LHS = RHS

5 0

2 years ago

Please help!!!!

DochEvi [55]

Answer:

256h28k8

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Y=3/2x+2 5x-y=5 At what point do they meet?

RSB [31]

(2,5) ...........................................................................................

7 0

3 years ago

The mean consumption of water per household in a city was 1425 cubic feet per month. Due to a water shortage because of a drough

liberstina [14]

Answer:

a) $t=\frac{1175-1425}{\frac{250}{\sqrt{100}}}=-10$

The degrees of freedom are given by:

$df=n-1=100-1=99$

The p value for this case would be given by:

$p_v =P(t_{99}$

Since the p value is significantly lower than he significance level given we have enough evidence to reject the null hypothesis and we can conclude that the true mean is lower than 1425

b) For this case we need to find a critical value in the t distribution with 99 degrees of freedom who accumulates 0.025 of the area in the right tail and we got:

$t_{\alpha/2}=1.984$

Since the calculated value is higher than the critical value we can reject the null hypothesis at the significance level provided and we can say that the true mean is higher than 1425

Step-by-step explanation:

Information given

$\bar X=1175$ represent the sample mean for the cubic feets of households

$\sigma=250$ represent the population standard deviation

$n=100$ sample size

$\mu_o =1425$ represent the value to verify

$\alpha=0.025$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Part a

We want to test that the mean consumption of water per household has decreased due to the campaign by the city council, the system of hypothesis would be:

Null hypothesis: $\mu \geq 1425$

Alternative hypothesis: $\mu < 1425$

Since we don't know the deviation the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

The statistic is given by:

$t=\frac{1175-1425}{\frac{250}{\sqrt{100}}}=-10$

The degrees of freedom are given by:

$df=n-1=100-1=99$

The p value for this case would be given by:

$p_v =P(t_{99}$

Since the p value is significantly lower than he significance level given we have enough evidence to reject the null hypothesis and we can conclude that the true mean is lower than 1425

Part b

For this case we need to find a critical value in the t distribution with 99 degrees of freedom who accumulates 0.025 of the area in the right tail and we got:

$t_{\alpha/2}=1.984$

Since the calculated value is higher than the critical value we can reject the null hypothesis at the significance level provided and we can say that the true mean is higher than 1425

6 0

3 years ago

In the arithmetic sequence, -12, -17, -22, -27... which term has a value of -68?​

In the arithmetic sequence, -12, -17, -22, -27... which term has a value of -68?