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RSB [31]
3 years ago
9

Estimate the length of tamara’s Room

Mathematics
1 answer:
andriy [413]3 years ago
6 0

Answer:

Step-by-step explanation:

What is the length of tamara's room?

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Find the discriminant of the following quadratic equation then state the number of rational, irrational, and imaginary solutions
lianna [129]

Answer:

  • -6x² - 6 = -7x - 9
  • -6x² + 7x - 6 + 9 = 0
  • -6x² + 7x + 3 = 0
  • 6x² - 7x - 3 = 0

<u>Discriminant:</u>

  • D = (-7)² - 4*6*(-3) = 49 + 72 = 121

<u>Since D > 0, there are 2 real solutions:</u>

  • x = (- (-7) ±√121 )/12
  • x = (7 ± 11)/12
  • x = 1.5, x = -1/3

8 0
3 years ago
Read 2 more answers
Tommy earned ​$ on Monday and ​$ on Tuesday. How many lucky bamboo plants can he buy with the total money he​ earned? lucky bamb
Montano1993 [528]

Answer: See explanation

Step-by-step explanation:

Your question isn't complete. But let's assume that the amount earned on Monday is $38 while $25 was earned on Tuesday.

The total amount earned for both days will be:

= $38 + $25

= $63

Since lucky bamboo plants are $7 each, we then divide the total amount gotten by $7. This will be:

= $63 / $7

= 9

Therefore, he can buy 9 lucky bamboos

7 0
2 years ago
Jim's cat can run 4 blocks in 15 seconds. Juanita says one of her cats can run 3 blocks in 11
vagabundo [1.1K]

Answer:

Juanita's cat is faster because it can run more blocks in a second

Step-by-step explanation:

Take the number of blocks and divide by the number of seconds

Jim:

4 blocks/ 15 seconds

.26666.... block per second

Juanita

3 blocks / 11 seconds

.2727...... blocks per second

Juanita's cat is faster because it can run more blocks in a second

5 0
2 years ago
I need the work to please
Free_Kalibri [48]
There you go I hope it helps

7 0
2 years ago
Read 2 more answers
An average computer mouse inspector can inspect 50 mice per hour. The 48 computer mice inspectors at a particular factory can on
Ronch [10]

Answer:  C. Yes, because -2.77 falls in the critical region .

Step-by-step explanation:

Let \mu be the population mean .

As per given , we have

H_0:\mu=50\\\\ H_a: \mu

Since the alternative hypothesis is left-tailed and population standard deviation is not given , so we need to perform a left-tailed t-test.

Test statistic : t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}

Also, it is given that ,

n= 48

\overline{x}=46

s= 10

t=\dfrac{46-50}{\dfrac{10}{\sqrt{48}}}=\dfrac{-4}{\dfrac{10}{6.93}}\\\\=\dfrac{-4}{1.443}\approx-2.77

Degree of freedom = df = n-1= 47

Using t-distribution , we have

Critical value =t_{\alpha,df}=t_{0.025,47}=2.0117

Since, the absolute t-value (|-2.77|=2.77) is greater than the critical value.

So , we reject the null hypothesis.

i.e. -2.77 falls in the critical region.

[Critical region is the region of values that associates with the rejection of the null hypothesis at a given probability level.]

Conclusion : We have sufficient evidence to support the claim that these inspectors are slower than average.

Hence, the correct answer is C. Yes, because -2.77 falls in the critical region

4 0
3 years ago
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