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Zina [86]
3 years ago
6

4. Transversal E F intersects AB and CD, as shown in

Mathematics
2 answers:
Zepler [3.9K]3 years ago
5 0

Answer: D, m<5=55.1 and m<7=124.9

Step-by-step explanation: 5 is the opposite of 4 in this diagram becuase of the bisector so that would make it the same 55.1, while 7 is an obtuse angle so it's clearly more than 90°, leaving 124.9 the most logical explanation.

Scorpion4ik [409]3 years ago
4 0
Ab cd ef ab cd ef ab ef ef ab
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What is the next letter a z b y c x d
Nataliya [291]
The next logical letter would be w.
This pattern starts at the first letter, then the last letter, then the second letter, then the second to last, the third, third to last, and so on.
6 0
4 years ago
Read 2 more answers
What is the solution to this equation: 5x to the 2 power - 36x - 81 = ?
Viefleur [7K]

Answer:

5x² - 36x - 81

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS

<u>Algebra I</u>

  • Combining Like Terms
  • Standard Form: ax² + bx + c = 0

Step-by-step explanation:

<u>Step 1: Define expression</u>

5x² - 36x - 81

<u>Step 2: Simplify</u>

<em>We cannot simplify the expression down further.</em>

7 0
3 years ago
PLEASE HELP ASAP!!
MA_775_DIABLO [31]
Yes, an additional 4% of Americans now support the President's policy on Afghanistan.

Is this the answer you needed?
8 0
3 years ago
If an object is projected upward with an initial velocity of 123 ft per​ sec, its height h after t seconds is h=−16t2+123t. Find
Alinara [238K]

Answer:

Height after 5 seconds is <em>215 ft</em>

Step-by-step explanation:

Given that:

The initial velocity of object which is projected upwards is 123 ft/sec.

Height of the object, <em>h</em> after time <em>t </em> in seconds,  is given as:

h=-16t^2+123t

Here, <em>t </em>will always be positive, so 123t and 16t^2 will also be positive.

But coefficient of 16t^2 is negative, that means something is subtracted from the positive term 123t.

To find:

Height of object after 5 seconds.

Solution:

Given that t=5 seconds.

Let us put the value in the given relation of h and t:

h=-16\times 5^2+123 \times 5\\\Rightarrow h=-400+615\\\Rightarrow \bold{h = 215\ ft}

So, height after 5 seconds is <em>215 ft</em>.

6 0
3 years ago
In a survey, 11 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped
pickupchik [31]

Answer:

The population standard deviation is not known.

90% Confidence interval by T₁₀-distribution: (38.3, 53.7).

Step-by-step explanation:

The "standard deviation" of $14 comes from a survey. In other words, the true population standard deviation is not known, and the $14 here is an estimate. Thus, find the confidence interval with the Student t-distribution. The sample size is 11. The degree of freedom is thus 11 - 1 = 10.

Start by finding 1/2 the width of this confidence interval. The confidence level of this interval is 90%. In other words, the area under the bell curve within this interval is 0.90. However, this curve is symmetric. As a result,

  • The area to the left of the lower end of the interval shall be 1/2 \cdot (1 - 0.90)= 0.05.
  • The area to the left of the upper end of the interval shall be 0.05 + 0.90 = 0.95.

Look up the t-score of the upper end on an inverse t-table. Focus on the entry with

  • a degree of freedom of 10, and
  • a cumulative probability of 0.95.

t \approx 1.812.

This value can also be found with technology.

The formula for 1/2 the width of a confidence interval where standard deviation is unknown (only an estimate) is:

\displaystyle t \cdot \frac{s_{n-1}}{\sqrt{n}},

where

  • t is the t-score at the upper end of the interval,
  • s_{n-1} is the unbiased estimate for the standard deviation, and
  • n is the sample size.

For this confidence interval:

  • t \approx 1.812,
  • s_{n-1} = 14, and
  • n = 11.

Hence the width of the 90% confidence interval is

\displaystyle 1.812 \times \frac{14}{\sqrt{10}} \approx 7.65.

The confidence interval is centered at the unbiased estimate of the population mean. The 90% confidence interval will be approximately:

(38.3, 53.7).

5 0
3 years ago
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