PLEASE HELP 20 POINTS IM CRYING SO HARD RN WILL GIVE BRAINLIEST :(((( ITS IMPOSSIBLE 10TH GRADE

2. Analyze this inequality.-4x < 12 a) x > -3 b) x < -3 c) x ≤ -3 d) -x < 3

lianna [129]

Answer:

answer is b

Step-by-step explanation:

-4x<12

x<12÷-4

x<-3

8 0

4 years ago

A rectangle is 2/5 inches long and 1/3 inches wide what is the area of the rectangle

Andrews [41]

Answer:

2/15 inches

Step-by-step explanation:

(2/5)*(1/3)=2/15in

6 0

3 years ago

Identify the quotient in the form a + bi. HELP ASAP PLEASE!!

Artist 52 [7]

Solving $\frac{4-5i}{2+3i}$ we get $-\frac{7}{13}-\frac{22i}{13}$

Option A is correct.

Step-by-step explanation:

We need to find the result of $\frac{4-5i}{2+3i}$

We need to remove i from the denominator. For removing i Multiplying and dividing by 2-3i:

$\frac{4-5i}{2+3i}*\frac{2-3i}{2-3i}$

Solving:

$=\frac{4-5i*2-3i}{2+3i*2-3i}\\=\frac{4(2-3i)-5i*(2-3i)}{(2)^2-(3i)^2}\\=\frac{8-12i-10i+15i^2)}{4-9i^2}\\We\,\,i^2=-1\\=\frac{8-12i-10i+15(-1))}{4-9(-1)}\\=\frac{8-12i-10i-15}{4+9}\\=\frac{8-15-12i-10i}{13}\\=\frac{-7-22i}{13}\\=-\frac{7}{13}-\frac{22i}{13}$

So, solving $\frac{4-5i}{2+3i}$ we get $-\frac{7}{13}-\frac{22i}{13}$

Option A is correct.

Keywords: Complex numbers

Learn more about Complex numbers at:

#learnwithBrainly

5 0

3 years ago

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. Whe

kherson [118]

Answer:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: