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3 years ago

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