Draw a number line from –2 to 2. Put tick marks every one-tenth unit. Graph a point with the given coordinate.

Find the Expression OF this equation.<br>c/3 / ac/4<br><br>QUESTION C

Fiesta28 [93]

Answer:

See below

Step-by-step explanation:

$(a) \: \: x = \frac{c}{3} \\ \\ \implies \: {x}^{2} = { \bigg( \frac{c}{3} \bigg) }^{2} \\ \\ \bold{\implies \: {x}^{2} = \frac{ {c}^{2} }{9} } \\ \\ (b) \: \: x + y = \frac{c}{3} + \frac{ac}{4} \\ \\\implies \: \bold{x + y = \frac{4c + 3ac}{12} } \\ \\ \frac{xy}{z} = \frac{ \frac{c}{3} \times \frac{ac}{4} }{ \frac{ {a}^{2} }{2c + 1} } \\ \\ = \frac{ \frac{ac ^{2} }{12} }{ \frac{ {a}^{2} }{2c + 1} } \\ \\ = \frac{a {c}^{2} }{12} \times \frac{2c + 1}{ {a}^{2} } \\ \\ \implies\bold{\frac{xy}{z} = \frac{ {c}^{2}(2c + 1) }{12a} }$

8 0

2 years ago

I neeeeeeddddd helppppppppppppppp

Flauer [41]

Answer:

The answer is x=9 :)

4 0

3 years ago

Read 2 more answers

20% of a is 11. What is a?

Reika [66]

Answer:

a=55

Step-by-step explanation:

This question is asking for you to find the whole (a), given a part (11). This can be done through multiplication. 20% is 1/5 of 100%. This means to find the whole number, you should multiply the part by 5. 11*5=55. So, a must equal 55.

We can check this answer by working backward. Do this by finding 20% of 55, and checking to see if it is 11. To find a percentage of a number, multiply that number by the percent in decimal form. So, 20% of 55 is equal to 0.2*55, which is 11. Therefore, the answer is correct.

5 0

2 years ago

Read 2 more answers

Find the x-intercepts (if any) for the graph of the quadratic function.

OleMash [197]

Here is the answer but basically the work you just plug in 6 for A, 12 for B, and 5 for C..into the quadratic formula and solve from there

8 0

3 years ago

Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature

Vesnalui [34]

Answer:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation

$\bar X=21.5$ represent the mean for the temperatures

$s=1.5$ represent the sample standard deviation

$n=40$ sample size

$\mu_o =22$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 22C, the system of hypothesis would be:

Null hypothesis: $\mu \geq 22$

Alternative hypothesis: $\mu < 22$

If we analyze the size for the sample is > 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

P-value

We can calculate the degrees of freedom like this:

$df=n-1=40-1=39$

Since is a one left tailed test the p value would be:

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

5 0

3 years ago