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jayda bought a baseball hat for $8 when it was originally $18 dollars. what percent discount did she get?

il63 [147K]

The percent discount was,

55.5556. It could be changed to 56%, but as a whole it would more than likely be 55%.

-Mabel <3

4 0

3 years ago

It is 3 p.m. in the Central Time Zone; therefore, it is what time in the Mountain Time Zone.

Aleksandr [31]

Central time is 1 hr ahead of mountain time.....so if central time is 3 p.m. then mountain time would be 2 p.m.

3 0

3 years ago

A school raised $4,111 in donations. They plan to use the money to buy new social studies textbooks. Each textbook costs $40. Es

Debora [2.8K]

Answer:

102 textbooks

Step-by-step explanation:

From the above question,

The total amount in donations = $4,111 . They plan to use the money to buy new social studies textbooks. Each textbook costs $40.

If :

$40 = 1 textbook

$4,111 = x text books

Cross Multiply

40 × x textbooks = $4,111 × 1

x = $4,111/40

x = 102.775 textbooks

Since the number of textbooks is whole number and it cannot in be decimal, the number of textbooks they can purchase is 102 textbooks

4 0

3 years ago

Match each expression to the description shown. Answers may be used more than once.

Aliun [14]

Ok so 2(x+1) would be two times the sum of a number and one. “Sum of a number and one” is just “x+1” and then it’s two times that because the two is outside of the parentheses.

And then 2x+1 would be the product of two and a number increased by 1. The product is two and a number is 2x and then increased by 1 is the +1

5 0

3 years ago

A sample of 41 observations is selected from one population with a population standard deviation of 3.4. The sample mean is 100.

Marizza181 [45]

Answer:

a) If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.

b) (-1.9886, 1.9886)

c) $t=\frac{(100 -98.4)-(0)}{\sqrt{\frac{3.4^2}{41}}+\frac{5.6^2}{45}}=1.617$

d) $p_v =2*P(t_{84}>1.617) =0.1096$

So with the p value obtained and using the significance level given $\alpha=0.05$ 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 5% of significance the mean of the group 1 is NOT significantly different than the mean for the group 2.

e) $p_v =2*P(t_{84}>1.617) =0.1096$

We can use the following excel code to calculate it: "=2*(1-T.DIST(1.617;84;TRUE))"

Step-by-step explanation:

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 = 0$

Alternative hypothesis: $\mu_1 -\mu_2\neq 0$

Our notation on this case :

$n_1 =41$ represent the sample size for group 1

$n_2 =45$ represent the sample size for group 2

$\bar X_1 =100$ represent the sample mean for the group 1

$\bar X_2 =98.4$ represent the sample mean for the group 2

$s_1=3.4$ represent the sample standard deviation for group 1

$s_2=5.6$ represent the sample standard deviation for group 2

Part a

If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.

Part b

On this case since the significance level is 0.05 and we are conducting a bilateral test we have two critical values, and we need on each tail of the distribution $\alpha/2 = 0.025$ of the area.

The distribution on this cas since we don't know the population deviation for both samples is the t distribution with $df=n_1+n_2 -2= 41+45-2=84$ degrees of freedom.

We can use the following excel codes in order to find the critical values:

"=T.INV(0.025,84)", "=T.INV(1-0.025,84)"

And we got: (-1.9886, 1.9886)

Part c

The statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}$

And now we can calculate the statistic:

$t=\frac{(100 -98.4)-(0)}{\sqrt{\frac{3.4^2}{41}}+\frac{5.6^2}{45}}=1.617$

The degrees of freedom are given by:

$df=41+45-2=84$

Part d

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

$p_v =2*P(t_{84}>1.617) =0.1096$

So with the p value obtained and using the significance level given $\alpha=0.05$ 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 5% of significance the mean of the group 1 is NOT significantly different than the mean for the group 2.

Part e

$p_v =2*P(t_{84}>1.617) =0.1096$

We can use the following excel code to calculate it: "=2*(1-T.DIST(1.617;84;TRUE))"

8 0

3 years ago