This scatter diagram shows the marks scored in a class test in English and Science

how does the appearence of a histogram change when u use many small intervals then when you use a few large intervals

vampirchik [111]

If you're using a few larger intervals, then your histogram looks more stocky. If you imagine drawing one, it's because you're adding more values into the same category which can make the difference between two intervals much more noticeable. If you're using smaller intervals, however, you can much more accurately assess the difference between two different intervals. For that reason, the transition between one and another interval would look much more 'fluid'.

5 0

3 years ago

7. Convert the subtraction problem into an addition problem. 10- (-5)

dalvyx [7]

Answer:

10 + 5

Step-by-step explanation:

If you are subtracting a negative then it become adding a positive. A way to remember this is because it has matching socks.

7 0

2 years ago

Which one is it and why? Please actual answers only

Andrej [43]

Answer:

The first answer, 3(pi) km^2

Step-by-step explanation:

3 x pi squared determines the area

7 0

2 years ago

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A college counselor is interested in estimating how many credits a student typically enrolls in each semester. The counselor dec

Ket [755]

Answer:

(a) The usual load is not 13 credits.

(b) The probability that a a student at this college takes 16 or more credits is 0.1093.

Step-by-step explanation:

According to the Central limit theorem, if a large sample (n ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.

The information provided is:

$Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18$

The sample size is, n = 100.

The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.

So,

$\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191$

(a)

The null hypothesis is:

H₀: The usual load is 13 credits, i.e. μ = 13.

Assume that the significance level of the test is, α = 0.05.

Construct a (1 - α) % confidence interval for population mean to check the claim.

The (1 - α) % confidence interval for population mean is given by:

$CI=\bar x\pm z_{\alpha/2}\times SE$

For 5% level of significance the two tailed critical value of z is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Construct the 95% confidence interval as follows:

$CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)$

As the null value, μ = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.

Thus, it can be concluded that the usual load is not 13 credits.

(b)

Compute the probability that a a student at this college takes 16 or more credits as follows:

$P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z$

Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.

3 0

2 years ago

Read 2 more answers

Please help me

iVinArrow [24]

Answer:

i think 4 th option

if u want to verify just substitute (4,-2)

8 0

3 years ago