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3 years ago

Choose exactly two answers that are correct.

Mathematics

2 answers:

Lera25 [3.4K]3 years ago

4 0

A and D are both correct answers

skelet666 [1.2K]3 years ago

4 0

A and d are correct it equals 640

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Frida applied the steps below to find the product of (–1.2)(–9.4).Step 1: (–1.2)(–9.4) = (–9.4)(–1.2)

Lemur [1.5K]

Answer:

Step-by-step explanation:

did the unit test

5 0

3 years ago

How do I factor trinomials of a form ax^2+bx+c when a=1 <br><br> In one paragraph

vitfil [10]

Answer:

Trinomials in the form $x^2+bx+c$ can often be factored as the product of two binomials.

Step-by-step explanation:

As we know that a polynomial with three terms is said to be a trinomial.

Considering the trinomial of a form

$ax^2+bx+c$

a = 1

$x^2+bx+c$

Trinomials in the form $x^2+bx+c$ can often be factored as the product of two binomials.

For example,

$x^2+7x+10$

$=\left(x^2+2x\right)+\left(5x+10\right)$

$=x\left(x+2\right)+5\left(x+2\right)$

$\mathrm{Factor\:out\:common\:term\:}x+2$

$=\left(x+2\right)\left(x+5\right)$

Therefore, Trinomials in the form $x^2+bx+c$ can often be factored as the product of two binomials.

6 0

2 years ago

Use the distributive property to find the price. 5($5.91) a. $30.45 b. $29.55 c. $25.50

enot [183]

5*5 = 25

5*9= 45

5*1= 5

25 dollars now add the cents

now it is $25.50

6 0

3 years ago

Read 2 more answers

Solve the inequality for x: 25 - 14x < 53*

Mamont248 [21]

Answer:

x must be greater than -53/14

(im not 100% sure but i think this is correct)

5 0

3 years ago

According to a study conducted by the Gallup Organization, the the proportion of Americans who are afraid to fly is 0.10. A rand

notsponge [240]

Answer: 0.1457

Step-by-step explanation:

Let p be the population proportion.

Given: The proportion of Americans who are afraid to fly is 0.10.

i.e. p= 0.10

Sample size : n= 1100

Sample proportion of Americans who are afraid to fly = $\hat{p}=\dfrac{121}{1100}=0.11$

We assume that the population is normally distributed

Now, the probability that the sample proportion is more than 0.11:

$P(\hat{p}>0.11)=P(\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}>\dfrac{0.11-0.10}{\sqrt{\dfrac{0.10(0.90)}{1100}}})\\\\=P(z>\dfrac{0.01}{0.0090453})\ \ \ [\because z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}} ]\\\\=P(z>1.1055)\\\\=1-P(z\leq1.055)\\\\=1-0.8543=0.1457\ \ \ [\text{using z-table}]$

Hence, the probability that the sample proportion is more than 0.11 = 0.1457

3 0

3 years ago