A gold chain with an original price of $150 is on sale for $87 what is the percent of the discount

can you afford to buy a $50 video game every month if you make $40 a week and spend an average of $150 a month on food and gasol

arsen [322]

Answer: The answer is false. Hope this helps :)

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Find the equation of the perpendicular bisector of line MN

marissa [1.9K]

Answer:

Y=(-1/2)x+9/2

Step-by-step explanation:

For straight lines y=mx+c

m=gradient

Find the gradient first-m=(y2-y1) /(x2-x1)

Take the points M and N and substitute in the equation

m=(5-2)/(-1-5)

m=-1/2

Gradient of line MN×gradient of perp line =-1

There fore gradient of perp line =2

3 0

2 years ago

Which pair of expressions are equivalent?

Amiraneli [1.4K]

Answer:

3n + 7 + n and 4n + 7

Step-by-step explanation:

4 0

4 years ago

mariarad [96]

Answer:

$p_v =P(z>2.087) =0.0184$

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the mean for the control group it's not significantly higher than the mean of the treatment group by 1 point.

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that

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

$\sigma=2.5$

And the statistic is given by this formula:

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

Where z follows a normal standard distribution

The system of hypothesis on this case are:

Null hypothesis: $\mu_c \leq \mu_t+1$

Alternative hypothesis: $\mu_c >\mu_t+1$

Or equivalently:

Null hypothesis: $\mu_c - \mu_t \leq 1$

Alternative hypothesis: $\mu_c-\mu_t>1$

Our notation on this case :

$n_c =45$ represent the sample size for group control

$n_t =45$ represent the sample size for group treatment

$\bar X_c =5.2$ represent the sample mean for the group control

$\bar X_t =3.1$ represent the sample mean for the group treatment

And now we can calculate the statistic:

$z=\frac{(5.2-3.1)-(1)}{2.5\sqrt{\frac{1}{45}}+\frac{1}{45}}=2.087$

And now we can calculate the p value using the alternative hypothesis:

$p_v =P(z>2.087) =0.0184$

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the mean for the control group it's not significantly higher than the mean of the treatment group by 1 point.

5 0

3 years ago

What is the degree of each monomial? 5x^7 a. 5 b. 35 c. 7 d. 12

Iteru [2.4K]

Add up all of the exponents

$5x^7$ has a degree of $\boxed{7}$

3 0

3 years ago