What's the answer please, I need help

Ella has 65 photo. Dan has 78 photos .who has more photos?explain

aev [14]

Answer:

Dan

Step-by-step explanation:

Ella has only 65 photos, while Dan has 78.

78 is a higher number than 65...

4 0

3 years ago

Please someone help

anygoal [31]

Answer:

Nerys pays 3.1/3.84,5=84,5£per year

Eleri pays 3.84,5=253,5£per year

253,5-84,5=169£ more pay Eleri

5 0

3 years ago

How to graph f(x)=(x-2)^2+5

tatyana61 [14]

That's a quadratic, a nice parabola in vertex form.

The parabola has a positive x^2 term, so it's a CUP, concave up positive. It will have a minimum at the vertex, which is (2,5). Plot that point.

Now we need a couple of guide points to draw the usual parabola going up from both sides of its vertex. We try x=0 giving (0,9) and see that x=4 also gives 9, (4,9). Plot the parabola through those two points and the vertex and you're done.

3 0

3 years ago

Read 2 more answers

What is the recursive formula for this sequence . 10,14,18,22,26,...

ad-work [718]

X+4 it is add 4 digit

6 0

3 years ago

All 423 Wisconsin public schools were all given a rating by the Wisconsin Department ofPublic Instruction based on several varia

Blizzard [7]

Answer:

$z = \frac{\bar X -\mu}{\sigma_{\bar X}}= \frac{70.9-71.5}{0.77}=-0.779$

From this result we can conclude that the value of 70.9 is 0.78 deviation below the true mean of 71.5 and that can be considered as unusual. If we conduct a hypothesis test or a confidence interval we will see that we have enough evidence to conclude that the true mean is not significantly different from 71.5.

Step-by-step explanation:

For this case from all the population we know that the population mean and deviation are:

$\mu = 71.5,\sigma = 4.87$

And we take a random sample of size n =40 and we got a sample mean calculated with the following formula:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}= \frac{\sum_{i=1}^{40} X_i}{40}=70.9$

And we want to test if this value is unusually low.

Since the sample size is large n>30 we can use the central limit theorem who says that the distribution for the sample mean is given by:

$\bar X \sim N (\mu , \frac{\sigma}{\sqrt{n}})$

And on this case if we replace the values that we have we got:

$\bar X \sim N (\mu_{\bar X}=71.5,\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}=\frac{4.87}{\sqrt{40}}=0.77)$

For this case we can calculate how many deviations above or below is our calculated value from the sample of size 40, using the z score given by:

$z = \frac{\bar X -\mu}{\sigma_{\bar X}}= \frac{70.9-71.5}{0.77}=-0.779$

From this result we can conclude that the value of 70.9 is 0.78 deviation below the true mean of 71.5 and that can be considered as unusual. If we conduct a hypothesis test or a confidence interval we will see that we have enough evidence to conclude that the true mean is not significantly different from 71.5.

3 0

3 years ago