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3 years ago

Seven tenths of a number plus fourteen is less than forty-nine

1 answer:

5 0

I'm going to assume you need this inequality solved.

First, write it as numbers, not words.

7/10n + 14 < 49

where "n" is the unknown number.

Second, if I were you, I'd change that fraction into a decimal, as it'll make life easier later on.

7/10 = 0.7

Now, solve it like you would any other equation.

0.7n + 14 < 49

0.7n +14 - 14 < 49 - 14

0.7n < 35

0.7n ÷ 0.7 < 35 ÷ 0.7

n < 50

The answer is n < 50

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Alika [10]

Answer:

y=-4(x-2)²+2

Step-by-step explanation:

===================

4 0

3 years ago

HELP ASAP PLEASE thanks youu

BlackZzzverrR [31]

Answer: Probably A

Step-by-step explanation: You need to use a ruler for this, because without one its impossible to get it exact

Using the ruler though, you'd measure each side of the wall and either use a ratio of 0.5in:4ft or multiply every 0.5in by 4ft.

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3 years ago

An environmentalist wants to find out the fraction of oil tankers that have spills each month. Step 2 of 2 : Suppose a sample of

rewona [7]

Answer:

The 80% confidence interval for the population proportion of oil tankers that have spills each month is (0.199, 0.257).

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 $1 - \frac{\alpha}{2}$ .

Suppose a sample of 333 tankers is drawn. Of these ships, 257 did not have spills.

333 - 257 = 76 have spills.

This means that $n = 333, \pi = \frac{76}{333} = 0.228$

80% confidence level

So $\alpha = 0.2$ , z is the value of Z that has a pvalue of $1 - \frac{0.2}{2} = 0.9$ , so $Z = 1.28$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.228 - 1.28\sqrt{\frac{0.228*0.772}{333}} = 0.199$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.228 + 1.28\sqrt{\frac{0.228*0.772}{333}} = 0.257$

The 80% confidence interval for the population proportion of oil tankers that have spills each month is (0.199, 0.257).

6 0

3 years ago

The desired percentage of Silicon Dioxide (SiO2) in a certain type of aluminous cement is 5.5. To test whether the true average

Reil [10]

Answer:

a) Null hypothesis : H₀ : μ = 5.5

Alternative Hypothesis : H₁ : μ < 5.5

b) The test statistic

|t| = |-3.33| = 3.33

c) P - value lies between in these intervals

0.001 < P < 0.005

Step-by-step explanation:

Step( i ):-

Given data the Population mean 'μ' = 5.5

The small sample size 'n' = 16

The sample mean (x⁻) = 5.25

Given the percentage of SiO2 in a sample is normally distributed with a sigma of 0.3.

Null hypothesis : H₀ : μ = 5.5

Alternative Hypothesis : H₁ : μ < 5.5

Level of significance ∝ = 0.01

Step(ii):-

The test statistic

$t = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }$

$t = \frac{5.25 -5.5}{\frac{0.3}{\sqrt{16} } }$

On calculation , we get

t = -3.33

|t| = |-3.33| = 3.33

Step(iii):-

P - value

The degrees of freedom γ = n-1 = 16-1 =15

The calculated value t = 3.33 (check t-table) lies between the 0.001 to 0.005

0.001 < P < 0.005

Condition(i)

P - value < ∝ then reject H₀

Condition(ii)

P - value > ∝ then Accept H₀

we observe that 0.001 < P < 0.005

P- value < 0.01

we rejected H₀

(or)

The tabulated value = 2.60 at 0.01 level of significance with '15' degrees of freedom

The calculated value t = 3.33 > 2.60 at 0.01 level of significance with '15' degrees of freedom

The null hypothesis is rejected

Conclusion:-

Accepted Alternative hypothesis H₁

The Claim that the true average is smaller than 5.5

4 0

3 years ago

Pls help It due asap

harkovskaia [24]

The equation for midpoint is

((x1+x2)/2 , (y1+y2)/2) = ((1+3)/2 , (-2+2)/2)
= ( 2,0)

3 0

3 years ago

Read 2 more answers