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3 years ago

I hope to better understand these math questions, could you help me?

Mathematics

1 answer:

mart [117]3 years ago

7 0

Answer is SSA (first choice)

cause LE perpendicular to BU so LEB = LEU = 90

hope it helps

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I plugged this into photo math, but apparently I have to pay to find out how to get the answer. can you guys help me out?

dimulka [17.4K]

Photo math is charging? We don't need it. We factor the difference of two squares.

$\dfrac{5t^2-500}{t+10} = \dfrac{5(t^2- 100)}{t+10} = \dfrac{5(t-10)(t+10)}{t+10} = 5(t-10) = 5t-50$

Answer: 5t-50

8 0

3 years ago

What is the quotient 36 divided by 6

Dafna11 [192]

Answer:

Step-by-step explanation:

36/6

6, 12,18,24,30,36

1 2. 3. 4. 5. 6

3 0

3 years ago

Find the value of the power

julsineya [31]

Answer:

Power is the zero at the end of the number like this 56 small 0 at the top right

Step-by-step explanation: Its not just the zero its whatever number is divided by the power it cant go past 50

3 0

2 years ago

What is the fully factored form of the expression 16x^2 - 49y^2

podryga [215]

Answer:

(4x + 7y)(4x - 7y)

Step-by-step explanation:

Rewrite 16 as 4^2

= 4^2x^2 - 49y^2

Rewrite 49 as 7^2

= 4^2x^2 - 7^2y^2

Apply the Exponent Rule Pt 1 ((a^m*b^m =(ab)^m))

= (4x)^2 - 7^2y^2

Apply the Exponent Rule Pt 2

= (4x)^2 - (7y)^2

Apply Difference of Squares Formula (( x^2-y^2 = (x + y)(x - y)

= (4x + 7y) (4x - 7y)

6 0

3 years ago

a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was

Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

Let $\bar X$ = sample average time spent studying

The z-score probability distribution for sample mean is given by;

Z = $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = population mean hours spent studying = 25 hours

$\sigma$ = standard deviation = 15 hours

n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < $\bar X$ < 30 hours)

P(29 hours < $\bar X$ < 30 hours) = P( $\bar X$ < 30 hours) - P( $\bar X$ $\leq$ 29 hours)

P( $\bar X$ < 30 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ < $\frac{ 30-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z < 2) = 0.97725

P( $\bar X$ $\leq$ 29 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ $\leq$ $\frac{ 29-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z $\leq$ 1.60) = 0.94520

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.

Therefore, P(29 hours < $\bar X$ < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

7 0

3 years ago