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kolbaska11 [484]
3 years ago
10

Please help me with this answer​

Mathematics
2 answers:
trasher [3.6K]3 years ago
8 0
I think it’s A cuz if u add normally A is the answer u get
boyakko [2]3 years ago
5 0

Answer: B. 2x^4+12x^2-8

Step-by-step explanation: x^2+4x+2 x 2x^2+3x-4

x^2 x 2x^2+4x+3x+2-4

2x^4+12x^2-8

I think its B because the question does say that you have to multiply.

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Ethan burns 650 calories when he runs for 1 hour suppose he runs 5 hours in one week.?
Ray Of Light [21]
1 hr=650
5x650=3250 calories in one week
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3 years ago
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Laura has 160 pounds of green beans at her vegetable stand. If she sells 10 pounds of green beans every hour, what is the percen
Mademuasel [1]

Given:

Laura has 160 pounds of green beans at her vegetable stand.

She sells 10 pounds of green beans every hour.

To find:

The percent decrease in green beans after 3 hours.

Solution:

We have,

Beans sold in 1 hour = 10 pounds

So, Beans sold in 3 hours = 30 pounds

Decrease in green beans after 3 hours is 30 pounds.

Now,

\%\text{Decrease}=\dfrac{\text{Decrease in green beans after 3 hours }}{\text{Initial amount of green beans}}\times 100

\%\text{Decrease in green beans after 3 hours}=\dfrac{30}{160}\times 100

\%\text{Decrease in green beans after 3 hours}=\dfrac{300}{16}

\%\text{Decrease in green beans after 3 hours}=18.75\%

Therefore, the correct option is A.

7 0
3 years ago
15 + 45 = 5( blank + 9)
Tatiana [17]

Answer:

Step-by-step explanation:

15 + 45 = 5( _ + 9)

60 = 5 ( 3 + 9)

60 = 5(12)

60 = 60

5 0
3 years ago
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Find the value of x.
IRINA_888 [86]

Answer:

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6 0
2 years ago
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In a survey of 1000 large corporations, 260 said that, given a choice between a job candidate who smokes and an equally qualifie
Usimov [2.4K]

Answer:

The 95% confidence interval for the proportion of corporations preferring a nonsmoking candidate is (0.2328, 0.2872)

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}.

For this problem, we have that:

n = 1000, p = \frac{260}{1000} = 0.26

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}} = 0.26 - 1.96\sqrt{\frac{0.26*0.74}{1000}} = 0.2328

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.26 + 1.96\sqrt{\frac{0.26*0.74}{1000}} = 0.2872

The 95% confidence interval for the proportion of corporations preferring a nonsmoking candidate is (0.2328, 0.2872)

4 0
3 years ago
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