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3 years ago

Please help me the question is on the file ;-;

Mathematics

2 answers:

Elenna [48]3 years ago

8 0

Answer:

The last answer.

Step-by-step explanation:

The area of a rectangle is

Area=(width)*(length)

If we substitute the 2 given values into the formula we will get

Area= 3(x+5)

=3x+15

Since it's asking that it has to be "at least 300", it's saying that the equation has to be 300 or more.

3x+15>(less than or equal to) 300.

12345 [234]3 years ago

7 0

Answer:

D) 3x + 15 ≥ 300

Step-by-step explanation:

The rectangle must have a area of AT LEAST 300 square ft, or 300 or greater. Use the greater than or equal sign: ≥ 300.

Note that your are solving for the area, and that this shape is a rectangle. To get the area of the rectangle, simply multiply length and width:

(3) * (x + 5) = Area of rectangle.

Distribute the 3 to all terms:

3(x + 5) = 3x + 15 = Area of rectangle.

Put the two parts together:

3x + 15 ≥ 300.

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Algebra 2A Help !<br><br> Tadakimasu ~

Ludmilka [50]

Function is p(x)=(x-4)^5(x^2-16)(x^2-5x+4)(x^3-64)

first factor into (x-r1)(x-r2)... form

p(x)=(x-4)^5(x-4)(x+4)(x-4)(x-1)(x-4)(x^2+4x+16)
group the like ones
p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)

multiplicity is how many times the root repeats in the function
for a root r₁, the root r₁ multiplicity 1 would be (x-r₁)^1, multility 2 would be (x-r₁)^2

so

p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)
(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the root 1 and has multiplity 1
(x^2+4x+16) is not on the real plane, but the roots are -2+2i√3 and -2-2i√3, each multiplicity 1 (but don't count them because they aren't real

baseically

(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the root 1 and has multiplity 1

7 0

2 years ago

Weinstein, McDermott, and Roediger report that students who were givne questions to be answered while studying new material had

emmainna [20.7K]

<h2>Answer with explanation:</h2>

Let $\mu$ be the population mean.

By considering the given information , we have

Null hypothesis : $H_0: \mu=73.4$

Alternative hypothesis : $H_a: \mu>73.4$

Since alternative hypothesis is right-tailed , so the test is a right-tailed test.

Given : Sample size : n=16 , which is a small sample , so we use t-test.

Sample mean: $\overline{x}=78.3$ ;

Standard deviation: $s=8.4$

Test statistic for population mean:

$t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$

i.e. $t=\dfrac{78.3-73.4}{\dfrac{8.4}{\sqrt{16}}}\approx2.333$

Using the standard normal distribution table of t , we have

Critical value for $\alpha=0.01$ : $t_{(n-1,\alpha)}=t_{(15,0.01)}=2.602$

Since , the absolute value of t (2.333) is smaller than the critical value of t (2.602) , it means we do not have sufficient evidence to reject the null hypothesis.

Hence, we conclude that we do not have enough evidence to support the claim that answering questions while studying produce significantly higher exam scores.

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2 years ago

Last Sunday, the average temperature was 8%, percent higher than the average temperature two Sundays ago. The average temperatur

MArishka [77]

Answer: 1.08T and (1+8/100)T

Step-by-step explanation:

Given: The average temperature two Sundays ago was T degrees Celsius.

Since, last Sunday, the average temperature was 8% higher than the average temperature two Sundays ago.

⇒ The average temperature last Sunday $=8\%\ of\ T+T$