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rewona [7]
3 years ago
11

Solve for m. m + 7 > 5

Mathematics
1 answer:
anastassius [24]3 years ago
5 0

Answer:Move all terms not containing  

m

to the right side of the equation.

Tap for more steps...

m

7

=

14

Multiply both sides of the equation by  

7

.

7

⋅

m

7

=

7

⋅

14

Simplify both sides of the equation.

Tap for more steps...

m

=

98

Step-by-step explanation:

pls give me  BRAINLIESTT PLSSSSSSS

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At the gym, Hillary swims every 6 days, runs every 4 days and cycles every 16 days. If she did all three activities today, What
Inga [223]

Step-by-step explanation:

We need to do LCM in this case.

2_|_ 6,4,16

2 _|_ 3,2,8

_|_ 3, 1,4

It is not divisible further.

So, LCM is 2×2×3×4 = 48

<u>So, 48 is the answer.</u>

Hope it helps :)

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8 0
4 years ago
Rosa brought d drawings to art show. After selling 15 of them, she had 38 left. Identify the equation that represents this situa
anyanavicka [17]

Answer:

23 minus 15 equals 38

Step-by-step explanation:

urwelcome

4 0
3 years ago
 Please help, Select the graph of the equation below. y=3/2x^2-6x.
valentinak56 [21]

So before we can decide which graph it is, we have to find the zeros (x-intercepts) of this graph. We can do this by setting y to 0.

Firstly, factor out 3x on the right side of the equation: 0=3x(\frac{x}{2}-2)

Next, use the zero product property to solve y = 0:

3x=0\\ x=0

\frac{x}{2}-2=0\\\\ \frac{x}{2}=2 \\ \\ x=4

So we know that the zeros of this equation are (4,0) and (0,0). Looking at the four graphs, the only graph that has a line crossing those 2 points is the first graph. Therefore, the graph of this equation is the first graph.

4 0
3 years ago
What is the square root of 98 round to the nearest tenth?
meriva
The square root of 98 is 9.899<span> so I think if it was rounded to the nearest tenth it would be 9.9</span>
4 0
3 years ago
Read 2 more answers
In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead. Construct a 95% conden
S_A_V [24]

Answer:

The 95% confidence interval for the proportion of water specimens that contain detectable levels of lead is (0.472,0.766).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead.

This means that n = 42, \pi = \frac{26}{42} = 0.619

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.619 - 1.96\sqrt{\frac{0.619*0.381}{42}} = 0.472

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.619 + 1.96\sqrt{\frac{0.619*0.381}{42}} = 0.766

The 95% confidence interval for the proportion of water specimens that contain detectable levels of lead is (0.472,0.766).

4 0
3 years ago
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