there are 15 cards numbered 1 through 15 . What is the probability that you choose a 5 or a 9? THANK YOU!!

Use the recursive formula to answer the question. What’s the 7th term in the sequence?

Answer:

a7 = 36

Step-by-step explanation:

Here, we are to use the recursive formula to calculate the 7th term( we add the preceding term to 4 so as to get the succeeding term)

a1 = 12

using the recursive formula given in the question

a2 = 12 + 4 = 16

a3 = 16 + 4 = 20

a4 = 20 + 4 = 24

a5 = 24 + 4 = 28

a6 = 28 + 4 = 32

a7 = 32 + 4 = 36

3 0

3 years ago

E = mc2 please help ive been looking and i can’t find an answer for math.

goldfiish [28.3K]

Answer:

Don't really understand the question, but

7 0

3 years ago

Is a rectangular prism has a length width and height of 3/8cm, 5/8cm and 7/8 in a meter respectively then the volume of the pris

myrzilka [38]

The volume of a rectangular prism is given by the multiplication of its three dimensions: if $w, h, l$ are the width, height and length of the prism, the volume is

$V = w\cdot h\cdot l$

So, in your case, you have

$V = \dfrac{3}{8} \cdot \dfrac{5}{8} \cdot \dfrac{7}{8} = \dfrac{3\cdot 5 \cdot 7}{8\cdot 8 \cdot 8} = \dfrac{105}{512}$

4 0

3 years ago

18.

ASHA 777 [7]

Answer:

19/100 which is already simplified( it can't be simplified)

How I got the answer: subtract 81 from 100 and you'll get 19.

Put 19 over 100 (19/100) and there's the answer.

3 0

3 years ago

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

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