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2 years ago

Help me please chile

Mathematics

2 answers:

jekas [21]2 years ago

7 0

Answer:

1 neither

2 supplementary

3 supplementary

Step-by-step explanation:

-Dominant- [34]2 years ago

7 0

Bejjjwjwnnaksuehvbsjsusue yes

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Divide. write each answer in simplest form.

Aleksandr [31]

Answer:

just do cross multiplication

Step-by-step explanation:

then ans will come

5 0

2 years ago

Read 2 more answers

Help anyone pleaseeee

AlexFokin [52]

Answer:

36 <127

(36,127°)

36 [ cos (127) + i sin (127)]

Step-by-step explanation:

w1 = 2 < 95

w2 = 18 < 32

w1 * w2

We multiply the magnitude and add the angles

w1 * w2 = 2*18 < (95+32)

=36 < 127

36 [ cos (127) + i sin (127)]

4 0

2 years ago

A University wanted to find out the percentage of students who felt comfortable reporting cheating by their fellow students. A s

Alla [95]

Answer:

The point estimate for this problem is 0.48.

Step-by-step explanation:

We are given that a University wanted to find out the percentage of students who felt comfortable reporting cheating by their fellow students.

A survey of 2,800 students was conducted and the students were asked if they felt comfortable reporting cheating by their fellow students. The results were 1,344 answered "Yes" and 1,456 answered "no".

Let $\hat p$ = proportion of students who felt comfortable reporting cheating by their fellow students

Now, point estimate ( $\hat p$ ) is calculated as;

$\hat p=\frac{X}{n}$

where, X = number of students who answered yes = 1,344

n = number of students surveyed = 2,800

So, Point estimate ( $\hat p$ ) = $\frac{1,344}{2,800}$

= 0.48 or 48%

5 0

3 years ago

What is the proportion in the first question?

Igoryamba

If two triangles are similar, the correspondant sides are in proportion. So, the proportion is

$15\div 10 = 18 \div 12 = x \div 6$

4 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

Read 2 more answers