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Varvara68 [4.7K]
3 years ago
12

I’ll give brainliest!!!

Mathematics
1 answer:
Maslowich3 years ago
6 0

14a. =

15b. 2/3 is less than 0.667

16c. 3 7/8 is greater than 3.375

17d. 3/8 is less than 1/2

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What is a factor of: tan^2x + 2tan(x)cos(x) + cos^2(x)
nikklg [1K]
Lets factor  this quadratic function 

the factors  are (tan x + cos x)(tan x + cos x)

So the answer is choice A.
6 0
3 years ago
The scale on the architectural plans for a new house is 1 in. equals 4 ft. Find the length and width (in ft) of a room that meas
Oksi-84 [34.3K]

Answer: the actual length and width of the room is 24ft by 36ft

Step-by-step explanation:

The scale on the architectural plans for a new house is 1 in. equals 4 ft. This means that every 1 inch on the on the architectural plan represents 4 feet on the actual building.

Therefore, if the length of the room as measured on the drawing is 6 inches, the actual length of the room will be

6 × 4 = 24 feet

Also, if the width of the room as measured on the drawing is 9 inches, the actual width of the room will be

9 × 4 = 36 feet

8 0
3 years ago
Select all the correct answers.
harkovskaia [24]

Answer:Which four inequalities can be used to find the solution to this absolute value inequality?3+2|x-11 <9

2(x - 1) <-3

x-1 <9

X-1 >-9

-2(x - 1) > 3

x-1-3

X-1 <3

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
A new sales training program has been instituted at a rent-to-own company. Prior to the training, 10 employees were tested on th
Lunna [17]

Answer:

C. H0: µD = 0, HA: µD <0

Step-by-step explanation:

We assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal. So for this case is better apply a paired t test.  

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.  

Let put some notation  

x=test value for pretest , y = test value for posttest  

x: 66, 94, 87, 84, 76, 88

y: 75, 100, 93, 85, 75, 90  

The system of hypothesis for this case are:  

Null hypothesis: \mu_D \geq 0  

Alternative hypothesis: \mu_D < 0  

Because the difference D is defined as Pretest-Postest. And we want to see if the postets score is higher than the pretest with th training.

The first step is calculate the difference d_i=x_i-y_i and we obtain this:  

d: -9, -6, -6, -1, 1, -2

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{-23}{6}=-3.833  

The third step would be calculate the standard deviation for the differences, and we got:  

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =3.763  

The 4 step is calculate the statistic given by :  

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-3.833 -0}{\frac{3.763}{\sqrt{6}}}=-2.494  

The next step is calculate the degrees of freedom given by:  

df=n-1=6-1=5  

Now we can calculate the p value, since we have a left tailed test the p value is given by:  

p_v =P(t_{(5)}  

The p value is lower than the significance level assumed \alpha=0.05, so then we can conclude that we can reject the null hypothesis. So we can say that the differences between Pretest and Posttest \mu_{pretest}-\mu_{postest} are less than 0.

3 0
3 years ago
Helppppp plzzzzzzz steppsss iff posiblleeee, will mark brailest!!!!!!!!!!!!!!!!!!!!!!!!!
dem82 [27]

Answer:

54

Step-by-step explanation:

126+x=180

x=180-126=54

x=54

6 0
3 years ago
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