Repeated Subtraction for 12/6

Does the rule y = 2 · x + 4 represent a linear or an exponential function? Explain.

Natali5045456 [20]

<span>B.
Linear function; the independent variable x is not an exponent.

But for me, answer is:
</span>
Linear function; <span>the function is in the form y = mx + b, where m = slope, b = y-intercept, and x is a real number.
</span>

5 0

3 years ago

kittens weigh about 100 grams when born and gain 7 to 15 grams per day. If a kitten weighs 100 grams at birth and gained 8 grams

scoray [572]

After 25 days. 300=triple 100+8x=300. is that 20 words?

8 0

3 years ago

Ellie drew ΔLMN, in which m∠LMN = 90°. She then drew ΔPQR, which was a dilation of ΔLMN by a scale factor of one half from the c

vaieri [72.5K]

Answer:

m∠P ≅ m∠L; this can be confirmed by translating point P to point L.

Step-by-step explanation:

Angle angle (AA) similarity postulate state that two triangles are similar if two of their corresponding angle is similar. The corresponding angle for each point of the triangles will be:

∠L=∠P

∠Q=∠M

∠N=∠R

Since the 2nd triangle made from dilation, it should maintain its orientation.

Option 1 is true, ∠P corresponds to ∠L. If you move/translate point P to point L, you can confirm it because their orientation is the same.

Option 2 is false, the triangle will be similar if ∠P=∠N but you can't confirm it with translation alone.

Option 3 and 4 definitely wrong because it speaking about length, not the angle.

6 0

3 years ago

Read 2 more answers

Convert 7-6ths into a mixed number

AlexFokin [52]

11/1.
Hope this helps,
and I hope you have a productive day!

8 0

3 years ago

Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunne

Pavlova-9 [17]

Answer:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

$p_v =P(t_{(15)}>1.507)=0.076$

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, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X_{CTS}=2.35$ represent the mean for the sample CTS

$\bar X_{N}=1.83$ represent the mean for the sample Normal

$s_{CTS}=0.88$ represent the sample standard deviation for the sample of CTS

$s_{N}=0.54$ represent the sample standard deviation for the sample of Normal

$n_{CTS}=10$ sample size selected for the CTS

$n_{N}=7$ sample size selected for the Normal

$\alpha=0.01$ 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 for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis: $\mu_{CTS} \leq \mu_{N}$

Alternative hypothesis: $\mu_{CTS} > \mu_{N}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

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

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{CTS}+n_{N}-2=10+7-2=15$

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

$p_v =P(t_{(15)}>1.507)=0.076$

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

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, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

6 0

4 years ago