Anyone can help asap?

Given that 4(n-5)+2=3(n-1),what is the value of n

Alina [70]

We'll first simplify both sides of the expression to get (simply using the distributive property):
$4n-20+2=3n-3$

Then, we can simplify a little bit:
$4n-18 = 3n-3$

To solve for n, we have to isolate it. Let's subtract 3n from both sides to further simplify it:
$n-18 = -3$

Then, add 18 to both sides to get the value of n:
$n = 15$

And that's your answer. If you have questions on how I got that, feel free to ask!

5 0

3 years ago

Read 2 more answers

Help please and thank u

Damm [24]

Answer:

Point S is your answer.

Step-by-step explanation:

See how all of them are rays, but they all connect at point S.

Point S is your answer.

6 0

3 years ago

Read 2 more answers

Which BEST shows where this function is only decreasing? A) −2 < x < 0.5 B) −∞ < x < −2 C) −2 < x < ∞ D) 0.5 &

Alexxx [7]

Answer:

in option B all the function is decreasing .thanks

8 0

3 years ago

Read 2 more answers

Based on an indication that mean daily car rental rates may be higher for Boston than for Dallas, a survey of eight car rental c

Elina [12.6K]

Answer:

$t=\frac{(47 -44)-(0)}{3\sqrt{\frac{1}{8}+\frac{1}{9}}}=2.058$

The degrees of freedom are:

$df=8+9-2=15$

And the p value would be:

$p_v =P(t_{15}>2.058) =0.0287$

Since we have a p value lower than the significance level given of 0.05 we can reject the null hypothesis and we can conclude that the true mean for car rental rates in Boston are signiﬁcantly higher than those in Dallas

Step-by-step explanation:

Data given

$n_1 =8$ represent the sample size for group Boston

$n_2 =9$ represent the sample size for group Dallas

$\bar X_1 =47$ represent the sample mean for the group Boston

$\bar X_2 =44$ represent the sample mean for the group Dallas

$s_1=3$ represent the sample standard deviation for group Boston

$s_2=3$ represent the sample standard deviation for group Dallas

We can assume that we have independent samples from two normal distributions with equal variances and that is:

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

Let the subindex 1 for Boston and 2 for Dallas we want to check the following hypothesis:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

The statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t student distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

Replacing we got:

$\S^2_p =\frac{(8-1)(3)^2 +(9 -1)(3)^2}{8 +9 -2}=9$

And the deviation would be just the square root of the variance:

$S_p=3$

The statitsic would be:

$t=\frac{(47 -44)-(0)}{3\sqrt{\frac{1}{8}+\frac{1}{9}}}=2.058$

The degrees of freedom are:

$df=8+9-2=15$

And the p value would be:

$p_v =P(t_{15}>2.058) =0.0287$

Since we have a p value lower than the significance level given of 0.05 we can reject the null hypothesis and we can conclude that the true mean for car rental rates in Boston are signiﬁcantly higher than those in Dallas

5 0

4 years ago

I NEED THE ANSWERS PLZ

g100num [7]

Answer:

1. 32

2. 4.7

3. 3

Step-by-step explanation:

4*8=32

6.5-1.8= 4.7

18/6=3 or 18 divided by 6 = 3

5 0

3 years ago

Read 2 more answers