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3 years ago

Sentence with enjoyment

Mathematics

2 answers:

Vanyuwa [196]3 years ago

5 0

I am delighted to fulfill your enjoyment of watching television.

Vera_Pavlovna [14]3 years ago

3 0

the state or process of taking pleasure in something.

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Full working out please ^^ for the first question , ty

sp2606 [1]

Answer:

Step-by-step explanation:

Given

y = 2x² - kx + 3

with a = 2, b = - k and c = 3

Since the curve touches the x- axis at one place then the roots are real and equal and the discriminant for this condition is

b² - 4ac = 0, that is

(- k)² - (4 × 2 × 3) = 0

k² - 24 = 0 ( add 24 to both sides )

k² = 24 ( take the square root of both sides )

k = ± $\sqrt{24}$ = ± 2 $\sqrt{6}$

k = 2 $\sqrt{6}$ or k = - 2 $\sqrt{6}$ → E

7 0

4 years ago

How can i use models to relate fractions and decimals?

Anit [1.1K]

You can use different colors

7 0

3 years ago

Read 2 more answers

Is 6200 ft greater less than 1 mile 900 ft

aliina [53]

The answer is 1 mile 900 feet.

5 0

4 years ago

Students in a representative sample of 69 second-year students selected from a large university in England participated in a stu

Serhud [2]

Answer:

95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].

Step-by-step explanation:

We are given that for the 69 second-year students in the study at the university, the sample mean procrastination score was 41.00 and the sample standard deviation was 6.89.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

P.Q. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean procrastination score = 41

s = sample standard deviation = 6.89

n = sample of students = 69

$\mu$ = population mean estimate

Here for constructing 95% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the true mean, $\mu$ is ;

P(-1.9973 < $t_6_8$ < 1.9973) = 0.95 {As the critical value of t at 68 degree

of freedom are -1.9973 & 1.9973 with P = 2.5%}

P(-1.9973 < $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ < 1.9973) = 0.95

P( $-1.9973 \times{\frac{s}{\sqrt{n} } }$ < ${\bar X -\mu}$ < $1.9973 \times{\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.9973 \times{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.9973 \times{\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ =[ $\bar X-1.9973 \times{\frac{s}{\sqrt{n} } }$ , $\bar X+1.9973 \times{\frac{s}{\sqrt{n} } }$ ]

= [ $41-1.9973 \times{\frac{6.89}{\sqrt{69} } }$ , $41+1.9973 \times{\frac{6.89}{\sqrt{69} } }$ ]

= [39.34 , 42.66]

Therefore, 95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].

5 0

3 years ago

Which is a counterexample that disproves the conjecture? After completing several multiplication problems, a student concludes t

Dahasolnce [82]

When you make the product of a binomial of the kind x + a times other binomial that is of the kind x - a, you obtain another binomial (not a trinomial), so any example with that form will be a counterexample that disproves the conjecture:

(x + a) * (x - a) = x^2 - a^2

For example, (x +3) * (x - 3) = x^2 - 9. So, not always the product of two binomials is a trinomial.

5 0

3 years ago