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ASHA 777 [7]
3 years ago
8

Kami has 7 liters of water to fill water bottles that each hold 2.8 liters. How many bottles can she fill?

Mathematics
2 answers:
erastovalidia [21]3 years ago
8 0

Answer:

2 and one half bottles can be filled.

Step-by-step explanation:

Kami has 7 liters of water to fill the water bottles.

Each bottle hold 2.8 liters of water.

We will divide 7 liters by 2.8 liters to get the number of bottles.

The number of bottles she can fill = \frac{7}{2.8}

                                                        = 2.5 bottles

2 and a half bottles can be filled.

Pani-rosa [81]3 years ago
6 0
Kami can fill 2.5 bottles.
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Find the roots of the equation x^2+2x+5=0
stiv31 [10]

Answer:

x = - 1 ± 2i

Step-by-step explanation:

we can use the discriminant b² - 4ac to determine the nature of the roots

• If b² - 4ac > , roots are real and distinct

• If b² - 4ac = 0, roots are real and equal

• If b² - ac < 0, roots are not real

for x² + 2x + 5 = 0

with a = 1, b = 2 and c = 5, then

b² - 4ac = 2² - (4 × 1 × 5 ) = 4 - 20 = - 16

since b² - 4ac < 0 there are 2 complex roots

using the quadratic formula to calculate the roots

x = ( - 2 ± \sqrt{-16} ) / 2

  = (- 2 ± 4i ) / 2 = - 1 ± 2i


7 0
3 years ago
Step 1 ax2 + bx + c = 0
butalik [34]

Answer:

step three

Step-by-step explanation:

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7 0
2 years ago
Difference of -150 and 325​
Montano1993 [528]

Step-by-step explanation:

- 150 - 325 =  - 340

3 0
2 years ago
Read 2 more answers
Assume the random variable X is normally distributed with mean mu equals 50 and standard deviation sigma equals 7. Compute the p
defon

Answer:

P(X>34) = 0.9889

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 50

Standard Deviation, σ = 7

We are given that the distribution of random variable X is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(X greater than 34)

P( X > 34) = P( z > \displaystyle\frac{34 - 50}{7}) = P(z > -2.2857)

= 1 - P(z \leq -2.2857)

Calculation the value from standard normal z table, we have,  

P(X>34) = 1 - 0.0111= 0.9889= 98.89\%

The attached image shows the normal curve.

4 0
3 years ago
Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. The
Tems11 [23]

Answer:

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

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 z-score that has a p-value of 1 - \frac{\alpha}{2}.

They randomly survey 387 drivers and find that 298 claim to always buckle up.

This means that n = 387, \pi = \frac{298}{387} = 0.77

84% confidence level

So \alpha = 0.16, z is the value of Z that has a p-value of 1 - \frac{0.16}{2} = 0.92, so Z = 1.405.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.74

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.8

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

6 0
3 years ago
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