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Arturiano [62]
3 years ago
9

Solve for a. -4a – 2a – 7 = 11 a = [?] Not a fraction answer

Mathematics
2 answers:
Brrunno [24]3 years ago
8 0

Answer:

sorry for point

plz

plz

plz

plz

zepelin [54]3 years ago
4 0

Answer:

Hello,

Answer: a=-3

Step-by-step explanation:

-4a-2a-7=11\\\\\Longleftrightarrow\ -6a-7=11\\\\\Longleftrightarrow\ 6a=-7-11\\\\\Longleftrightarrow\ 6a=-18\\\\\Longleftrightarrow\ a=-3\\

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Bus A and Bus B leave the bus depot at 8 am. Bus A takes 25 minutes to do its route and bus B takes 35 minutes to complete its r
Dafna1 [17]

bus a: 25 + 25= 50 minutes x 12 = 6:00

bus b: 35 +35= 70 minutes x 12 = 8:40

3 0
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Free Promo Moneybag Yo - Fire?
alexandr402 [8]

Answer:

thxssss

Step-by-step explanation:

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8 0
3 years ago
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George went to school for 216 days in a full year. If his attendance in 60%, find the number of days on which the school was ope
inessss [21]
Given:
days present = 216 days
percentage of presence = 60%

Find the total number of days the school was opened.

216 days is the part of a whole. 60% is the percentage of the part to the whole.

To get the whole, we need to divide the part by its percentage

216 ÷ 60% = 360 days

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8 0
3 years ago
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A sprinkler system is being installed in a newly renovated building on campus. The average activation time is supposed to be at
melomori [17]

Answer:

A) t = 1.73

B) p-value =  0.0558

C) Data is not statistically significant because the p-value of 0.0558 is more than the significance value of 0.05

D) The decision means that the design specifications are not met.

E) Type II error

Step-by-step explanation:

The hypotheses are:

H₀: μ = 20

H₁: μ > 20

A) Formula for the test statistic is;

t = (x' - μ)/(s/√n)

Now, we are given;

x' = 21.5

μ = 20

s = 3

n = 12

Thus;

t = (21.5 - 20)/(3/√12)

t = 1.73

B) we have our t-value as 1.73

Now, Degree of freedom(DF) = n - 1

So,DF = 12 - 1 = 11

Using significance level of α = 0.05, t-value = 1.73 and DF = 11, one tailed hypothesis, from online P-value calculator attached, we have;

p-value =  0.0558

C) Data is not statistically significant because the p-value of 0.0558 is more than the significance value of 0.05

D) We will not reject the null hypothesis. The decision means that the design specifications are not met.

E) If the true average activation time of the sprinkler system is, in fact, equal to 20 seconds, then the null hypothesis is false.

Since we did not reject the null hypothesis even though it is false, the error that was committed was therefore a type II error.

4 0
3 years ago
Find area of blue shaded region
RUDIKE [14]

Answer:

\approx \: 7.33 \:  {cm}^{2}

Step-by-step explanation:

Area of the blue shaded region

=  \frac{ \theta}{360 \degree}  \times \pi r_1^2  - \frac{ \theta}{360 \degree}  \times \pi r_2^2  \\  \\  =   \frac{ \pi\theta}{360 \degree} ( r_1^2 - r_2^2) \\  \\ ( r_1 = 4 \: cm, \:  \:  r_2 = 3\: cm, \:  \:  \theta = 120 \degree) \\  \\ =   \frac{ 3.14 \times 120 \degree}{360 \degree} ( 4^2 - 3^2) \\  \\  =   \frac{ 3.14 }{3}  \times 7 \\  \\  =  \frac{21.98}{3}  \\  \\  = 7.32666667 \\  \\  \approx \: 7.33 \:  {cm}^{2}

8 0
3 years ago
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