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3 years ago

Please give answer ofHCF of 15 and 18

1 answer:

6 0

Answer: The prime factorization of 15 is: 3 x 5. The prime factorization of 18 is: 2 x 3 x 3. The prime factors and multiplicities 15 and 18 have in common are: 3. 3 is the gcf of 15 and 18.

Step-by-step explanation:

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The rectangular prism has a base that is 3 \times 9\text{ m}3×9 m3, times, 9, start text, space, m, end text and a height of 2\t

lawyer [7]

Answer:

Volume of green house = 54 m³

Step-by-step explanation:

Given:

Base = 3m x 9m

Height of house = 2 m

Find:

Volume of green house

Computation:

Volume = lbh

Volume of green house = 3m x 9m x 2m

Volume of green house = 54 m³

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3 years ago

Graph the image of F(-8, 10) after a rotation 180° counterclockwise around the origin.

pychu [463]

Answer:

F' is at (8, -10)

Step-by-step explanation:

clockwise or counter-clockwise do not matter when the rotation is 180°

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3 years ago

There are 130 people in a sport centre. 73 people use the gym. 62 people use the swimming pool. 58 people use the track. 22 peop

Maru [420]

p(the person does not use any facility)= 59/130

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3 years ago

In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s

Oksi-84 [34.3K]

Answer:

a) $\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

b) From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

c) $P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

$P(Z\geq2.070)=1-P(Z$

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$

And we can use the z score defined as:

$z=\frac{\bar x -\mu}{\sigma_{\bar x}}$

And using this we got:

$P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

And using a calculator, excel or the normal standard table we have that:

$P(Z\geq2.070)=1-P(Z$

8 0

3 years ago

Need help with questions 9,10 and 11 from the bottom<br><br> Will give brainliest!

qwelly [4]

Step-by-step explanation:

A=V/t

A=8.0m/s•1/2.0s

A=8.0m/2.0s²

A=4.0m/s²

10:

V=A•time

V=4.0m/s²•2.3s

V=9.2m/s

11:

V=A•t

24.0=5.6t

4.3s=t

5 0

4 years ago

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Please give answer ofHCF of 15 and 18​

Please give answer ofHCF of 15 and 18