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2 years ago

What is the greatest common factor of 28 and 46?

Mathematics

2 answers:

tester [92]2 years ago

8 0

28=2*2*7=2²*7; 46=2*23=>the greatest common factor is 2

Aloiza [94]2 years ago

3 0

2 is the only number that goes into both. Cheers!

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In the right triangle shown in the diagram below,

adelina 88 [10]

QUESTION :-

what is the value of x to the nearest whole number?

ANSWER :-

$\cos( \theta) = \frac{base}{hypotenuse} \\ \cos(30) = \frac{x}{24} \\ \frac{ \sqrt{3} }{ \cancel2} = \frac{x}{ \cancel{24}} = \frac{x}{12} \\ 1.732 = \frac{x}{12} \\ 12 \times 1.732 = x \\ x = 20.7$

the nearest whole number-->

20.7 --> approximately --> 21 units

5 0

2 years ago

Can you find the rate of change for the following function: -6x + 3y =9?

IgorC [24]

Answer:

The general rate of change can be found by using the difference quotient formula. To find the average rate of change over an interval, enter a function with an interval:

f ( x ) = x ^2 , [ 2 , 3 ]

The rate of Change is 2

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Use the formula for volume = ℎ to find the<br> volume of the rectangular prism shown.

Sedaia [141]

Answer: 56 / 729

Step-by-step explanation: (4 * 2 * 7) / (9^3)

3 0

2 years ago

Question is on picture<br><br> Please help<br> Zoom in if needed :D

Anastasy [175]

96$ source: just trust me bro

8 0

2 years ago

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