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sergiy2304 [10]
4 years ago
12

Zach is taller than Matt and Richard is shorter than Zach. Which of the following statements would be most accurate?

Mathematics
2 answers:
Nikitich [7]4 years ago
7 0
C) Richard is as tall as Matt
Lostsunrise [7]4 years ago
5 0

Answer:

B, Richard is shorter than Matt

Step-by-step explanation:

if Zach is taller than Matt then point E is eliminated, if Richard is shorter than Zach then points A, C and E are out of the question because if Zach is shorter than Matt then there is no way that he is taller than Zach, and point D is eliminated.

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a 400 g of a liquid of density 1.6g/cm3 is made from mixing liquids of density 1.2g/cm3 and 1.8g/cm3 find the mass of the liquid
gogolik [260]

Answer:

133.33g

Step-by-step explanation:

Let the:

Mass of 1.2g/cm³ of liquid = x

Mass of 1.8g/cm³ of liquid = y

From our Question above, our system of equations is given as:

x + y = 400........ Equation 1

x = 400 - y

1.2 × x + 1.8 × y = 1.6 × 400

1.2x + 1.8y = 640..... Equation 2

We substitute, 400 - y for x in Equation 2

1.2(400 - y) + 1.8y = 640

480 - 1.2y + 1.8y = 640

- 1.2y + 1.8y = 640 - 480

0.6y = 160

y = 160/0.6y

y = 266.67 g

Solving for x

x = 400 - y

x = 400 - 266.67g

x = 133.33g

Therefore, the mass of the liquid of density 1.2g/cm³ is 133.33g

7 0
3 years ago
Keisha walked 12 miles in 3 hours. Which rate represents this situation?
GaryK [48]
If you re-read your question, you will see that the answer you want is the one that includes 12 miles and 3 hours which is only choice A
7 0
4 years ago
Read 2 more answers
Thank you to whoever answers! :))
Elena L [17]

Answer:

C. Points L,J and K are collinear.

Step-by-step explanation:

Collinear points are points that lie on the same line. So in the picture only the points L,J and K are collinear.

8 0
2 years ago
Read 2 more answers
How to solve question 19<br> thanks
miskamm [114]
Let X be the random variable for the weight of any given can, and let \mu and \sigma be the mean and standard deviation, respectively, for the distribution of X.

You have

\begin{cases}\mathbb P(X377)=0.025\end{cases}

Recall that for any normal distribution, approximately 99.7% of it lies within three standard deviations of the mean, i.e. \mathbb P(\mu-3\sigma. This means 0.3% must lie outside this range, \mathbb P(X\mu+3\sigma)=0.003. Because the distribution is symmetric, it follows that \mathbb P(X.

Also recall that for any normal distribution, about 95% of it falls within two standard deviations of the mean, so \mathbb P(\mu-2\sigma, which means 5% falls outside, and by symmetry, \mathbb P(X>\mu+2\sigma)=0.025.

Together this means

\begin{cases}\mu-3\sigma=347\\\mu+2\sigma=377\end{cases}

Solving for the mean and standard deviation gives \mu=365 and \sigma=6.
6 0
4 years ago
An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on averag
shtirl [24]

Answer: No, the data does not support the claim at 1% level as the mean time is no longer than 4.5 years.

Step-by-step explanation:

Since we have given that

n the California state university system take 4.5 years, on average, to finish their undergraduate degrees.

So, the hypothesis would be

H_0:\mu=4.5\\\\H_a:\mu>4.5

Mean = 5.1

Standard deviation = 1.2

n = 49

So, test statistic value would be

z=\dfrac{\bar{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\z=\dfrac{4.5-5.1}{\dfrac{1.2}{\sqrt{49}}}\\\\z=\dfrac{-0.6}{0.17}\\\\z=-3.53

At 1% level of significance, critical value is 2.58

Since 2.58>-3.53.

So, we will accept the null hypothesis.

Hence, No, the data does not support the claim at 1% level as the mean time is no longer than 4.5 years.

8 0
3 years ago
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