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3 years ago

1/4(1/3k+9)=6 step 1- 1/3k=15 step 2- 1/3k=15 step 3- k=5 find Olgas mistake step 1 step 2 steps 3 Olga did not make a mistake

Mathematics

1 answer:

shtirl [24]3 years ago

3 0

Answer:

Step 3 contains error.

Step-by-step explanation:

The given equation is :

$\dfrac{1}{4}(\dfrac{1}{3}k+9)=6$

Step 1.

Cross multiplying,

$(\dfrac{1}{3}k+9)=24$

Step 2.

Subtract 9 from both sides

$\dfrac{1}{3}k+9-9=24-9\\\\\dfrac{1}{3}k=15$

Step 3.

Cross multiplying

$k=15\times 3\\\\k=45$

So, there is an error in step 3. The correct answer should be 45.

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Indicate the method you would use to prove the two ▲'s ≅. If no method applies, enter "none".

zysi [14]

Answer: None

Step-by-step explanation:

In the given picture , there are two triangles and their corresponding angles are equal.

So by AAA similarity criteria they are similar. But there is no other information is given to prove that they are congruent.

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3 years ago

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Please help me with this

Eduardwww [97]

Answer:

$r^{6}$

Step-by-step explanation:

Using the rule of exponents

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3 years ago

The function y=-16t^2 +36 represents the height y (in feet) of a water droplet t seconds after falling from an icicle. After how

NikAS [45]

Answer:

1.5 seconds

Step-by-step explanation:

plug 0 in for y

0=-16t^2+36

then slove for t

-36=-16t^2

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9/4=t^2

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2 years ago

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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.

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3 years ago

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