Name two pairs of adjacent angles and two pairs of vertical angles in the figure

which expression is equivalent to 100 n2 − 1? (10n)2 − (1)2 (10n2)2 − (1)2 (50n)2 − (1)2 (50n2)2 − (1)2

Wewaii [24]

You know that $100=10^2$ .

Use the property $(ab)^2=a^2\cdot b^2$ to state that

$100n^2=10^2\cdot n^2=(10n)^2.$

Then the expression $100n^2-1$ is equivalent to

$(10n)^2-1=(10n)^2-1^2.$

Answer: correct choice is A.

5 0

3 years ago

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If each cookie contains 3.7 grams of fat, how much fat will 5 cookies have?

Mamont248 [21]

5 cookies will have 18.5 grams of fat.

8 0

3 years ago

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Find the slope of the line passing through the points (9,7) and (9, -2)

Semmy [17]

Slope = 9
i did the same thing

3 0

2 years ago

Write and evaluate the expression. Then, complete the statements.

Leni [432]

Answer:

The Answer is 5

Step-by-step explanation:

First, write the expression

30/t – 10

.

Second,

substitute

2 in for the variable, t.

Third,

simplify

by using

order of operations

.

The answer is

5

.

4 0

3 years ago

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A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classe

stealth61 [152]

Answer:

$t=\frac{(4-3.5)-0}{\sqrt{\frac{1.5^2}{11}+\frac{1^2}{9}}}}=0.890$

The p value for this case would be:

$p_v =P(t_{18}>0.890)=0.193$

The p value is higher than the significance level so then we can conclude that we can FAIL to reject the null hypothesis and then the true mean for group A is not significantly higher than the mean for B

Step-by-step explanation:

Information given

$\bar X_{1}=4$ represent the mean for sample A

$\bar X_{2}=3.5$ represent the mean for sample B

$s_{1}=1.5$ represent the sample standard deviation for A

$s_{2}=1$ represent the sample standard deviation for B

$n_{1}=11$ sample size for the group A

$n_{2}=9$ sample size for the group B

$\alpha=0.1$ Significance level provided

t would represent the statistic

Hypothesis to test

We want to verify if the student who graduates from college A has taken more math classes, on the average, the system of hypothesis would be:

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

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

The statistic is given by:

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

And the degrees of freedom are given by $df=n_1 +n_2 -2=11+9-2=18$

Replacing we got:

$t=\frac{(4-3.5)-0}{\sqrt{\frac{1.5^2}{11}+\frac{1^2}{9}}}}=0.890$

The p value for this case would be:

$p_v =P(t_{18}>0.890)=0.193$

The p value is higher than the significance level so then we can conclude that we can FAIL to reject the null hypothesis and then the true mean for group A is not significantly higher than the mean for B

5 0

3 years ago