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2 years ago

For what value of "a" will the product shown below have a purely imaginary value? (Show work please)

1 answer:

5 0

If you expand out the brackets you get this,

(4+5i)(a+2i) = 4a + (5a)i + 8i - 10

The -10 comes from 5i * 2i.
Squaring i becomes -1.

Let's group the real stuff together,
and imaginary separately,

(4a - 10) + (5a + 8)i

For this to be purely imaginary,
the real part needs to be zero.

Therefore 4a - 10 = 0

Solve for a.

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Help me ASAP!

butalik [34]

Answer:

the answer is b

Step-by-step explanation:

There is no negative sign therefore d and a are not an option and yep ito b

5 0

1 year ago

Which of the following is a valid probability distribution? A 2-column table labeled Probability Distribution A has 4 rows. The

dsp73

Answer:

I believe it is A

Step-by-step explanation:

all variables should add up to 1.00 and all in the first answer add up to 1.00

8 0

2 years ago

- If 2x2 + 8y = 121.5 and x2 - 8y = 121.5, then x =<br> 9<br> 6.36<br> 16<br> 7

Natalija [7]

Step-by-step explanation:

2x² + 8y = 121.5

x² - 8y = 121.5

Add the equations together to eliminate the y terms.

3x² = 243

x² = 81

x = 9

4 0

2 years ago

Estimate the value 9.9 squared times 1.79 and the square root of 97.5 divided by 1.96

Inessa05 [86]

Answer:

200 and 5

Step-by-step explanation:

1. 9.9 rounded is 10

1.79 rounded is 2

10^2 is 100

100 x 2 = 200

2. 97.5 rounded is 100

1.96 rounded is 2

square root of 100 is 10

10/2 = 5

8 0

2 years ago

A recent study reported that high school students spend an average of 94 minutes per day texting. Jenna claims that the average

larisa [96]

Answer:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

Step-by-step explanation:

Jenna claims that the average time of texting at her larger high school is greater than 94 minutes per day.

From here we can see that we have to perform a hypothesis test about a sample mean. The null and alternate hypothesis will be:

Null Hypothesis: $\mu \leq 94$

Alternate Hypothesis: $\mu > 94$

Jenna collected data from a sample of 32 students. So, sample size will be:

Sample Size = n = 32

Sample Mean = x = 96.5

Sample Standard Deviation = s = 6.3

We have to perform a hypothesis test, to test Jenna's claim. Since, the value of Population Standard Deviation is unknown and the value of Sample Standard Deviation is known, we will use One Sample t-test in this case.

The formula to calculate the test statistic is:

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

Using the values, we get:

$t=\frac{96.5-94}{\frac{6.3}{\sqrt{32} } }=2.245$

The degrees of freedom will be:

df = n - 1 = 32 - 1 = 31

We have to convert the t-score 2.245 with 31 degrees of freedom to its equivalent p-value. From t-table this value comes out to be:

p-value = 0.0160

The significance level is:

$\alpha =0.05$

Since, the p-value is lesser than the level of significance, we reject the Null Hypothesis.

Conclusion:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

3 0

2 years ago