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ss7ja [257]
2 years ago
15

Please help me I need the answer for this question. correct one please

Mathematics
1 answer:
andreyandreev [35.5K]2 years ago
7 0

Answer:

B

Step-by-step explanation:

Because to find out if two triangles are similar you need an ASA which can help determine if they are similar or not.

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QUESTION 1 Name the type of transformation depicted by the picture. reflection rotation translation none of the above
nydimaria [60]
I would be happy to help but it seems there are no pictures attached to answer from. 
7 0
3 years ago
A triangle has a base of 4x-7 and a height of 2x + 3. Find the area.
wolverine [178]

Answer:

area = ‐11

Step-by-step explanation:

hope this is correct and helps

7 0
3 years ago
Will give brainlist to first correct answer----The given data set shows the number of books read by eight children in a reading
Lesechka [4]
40 is considered an outlier since it it much further out on the data scale
Hope this helps!
8 0
3 years ago
If the mean is 298 and standard deviation is 22 what is the probability that a random sample of 42 pregnancies has a mean gestat
anyanavicka [17]

Answer:

The probability is 0.9909.

Step-by-step explanation:

Test statistic (z) = (sample mean - population mean) ÷ (sd/√n)

sample mean = 290 days

population mean = 298 days

sd = 22 days

n = 42

z = (290 - 298) ÷ (22/√42) = -8 ÷ 3.395 = -2.36

The cumulative area of the test statistic is the probability that the mean gestation period is less than 290 days. The cumulative area is 0.9909. Therefore the probability is 0.9909.

7 0
3 years ago
Read 2 more answers
The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50, with a standard deviation of $5.00. Assu
m_a_m_a [10]

Answer: 0.0170

Step-by-step explanation:

Given : The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50, with a standard deviation of $5.00.

i.e. \mu=23.50

\sigma=5

We assume the distribution of amounts purchased follows the normal distribution.

Sample size : n=50

Let \overline{x} be the sample mean.

Formula : z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

Then, the probability that the sample mean is at least $25.00 will be :-

P(\overline{x}\geq\25.00)=P(\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\geq\dfrac{25-23.50}{\dfrac{5}{\sqrt{50}}})\\\\=P(z\geq2.12)\\\\=1-P(z

Hence, the likelihood the sample mean is at least $25.00= 0.0170

5 0
3 years ago
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