Write a fraction less than 1, with a demoninator of 6 that is greater than 3/4

Solve for a a^2+17=42?

grandymaker [24]

We are given the following equation:

$a^2+17=42$

Subtract both sides by 17

$a^2=25$

Take the positive/negative square root of both sides

$a=\pm5$

This should be your answer. Let me know if you have any questions, thanks!

6 0

3 years ago

Luis and Berto sell TVs. Last month berto sold 15 more TVs. Together they sold 163. How many TVs did Luis sell?

Juli2301 [7.4K]

Luis sold 148 TVs.

Step-by-step explanation:

163-15=148

3 0

3 years ago

Translate this into an equation <br> Seven times the sum of a number and 9 equals 8.

zhenek [66]

Answer:

7(9+n)=8

Step-by-step explanation:

I hope this is right :)))

5 0

3 years ago

Activity ||| : Define the following terms:

slamgirl [31]

7).Venn diagram is an illustration that uses circles to show the relationships among things or finite groups of things. Circles that overlap have a commonality while circles that do not overlap do not share those traits. Venn diagrams help to visually represent the similarities and differences between two concepts.

8).In set theory, the union of a collection of sets is the set of all elements in the collection. It is one of the fundamental operations through which sets can be combined and related to each other. A nullary union refers to a union of zero sets and it is by definition equal to the empty set.

9).The intersection of two or more given sets is the set of elements that are common to each of the given sets. The intersection of sets is denoted by the symbol '∩'. In the case of independent events, we generally use the multiplication rule, P(A ∩ B) = P( A )P( B ).

10).Sets in mathematics, are simply a collection of distinct objects forming a group. A set can have any group of items, be it a collection of numbers, days of a week, types of vehicles, and so on. Every item in the set is called an element of the set. ... Sets are usually represented using a roster form or a set builder form.

Step-by-step explanation:

please mark me brainliest

8 0

3 years ago

A student at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the Un

WINSTONCH [101]

Answer:

Part 1: The statistic

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022 \approx 0.00002$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

Step-by-step explanation:

Data given

$\bar X_{1}=5135$ represent the mean for four year college

$\bar X_{2}=4436$ represent the mean for two year college

$s_{1}=783$ represent the sample standard deviation for four year college

$s_{2}=553$ represent the sample standard deviation two year college

$n_{1}=35$ sample size for the group four year college

$n_{2}=35$ sample size for the group two year college

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to check if the mean enrollment at four-year colleges is higher than at two-year colleges in the United States , the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We can assume that the normal distribution is assumed since we have a large sample size for each case n>30. So then the sample mean can be assumed as normally distributed.

Part 1: The statistic

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

6 0

3 years ago