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Nostrana [21]
3 years ago
11

Becca is cooking lentil soup for a large dinner.

Mathematics
2 answers:
natima [27]3 years ago
5 0

Answer:

IT is B

Step-by-step explanation:

slamgirl [31]3 years ago
4 0

Answer:

Answer is B

Step-by-step explanation:

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HELP FOR 15 POINTS!The table shows the number of minutes in different numbers of days.
GalinKa [24]
B is the right answer m=1440d
6 0
4 years ago
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Simplify (3x2 - 2) + (2x2 - 6x + 3). (1 point)
Oxana [17]

Answer:

A) 5x² - 6x + 1

Step-by-step explanation:

In this question, you would have to simplify the expression.

Solve:

(3x² - 2) + (2x² - 6x + 3)

Combine like terms.

3x² - 2 + 2x² - 6x + 3)

5x² - 2 - 6x + 3

5x² - 6x + 1

Since there's no more like terms, we have completely simplified the expression.

Therefore, your answer would be A) 5x² - 6x + 1

6 0
3 years ago
BAD is congruent to BCD by <br> the
scoundrel [369]
AB is congruent to BC
6 0
4 years ago
Each day a commuter takes a bus to work, the transportation system has a phone app that tells her what time the bus will arrive.
Paraphin [41]

Answer:

Step-by-step explanation:

Hello!

The commuter is interested in testing if the arrival time showed in the phone app is the same, or similar to the arrival time in real life.

For this, she piked 24 random times for 6 weeks and measured the difference between the actual arrival time and the app estimated time.

The established variable has a normal distribution with a standard deviation of σ= 2 min.

From the taken sample an average time difference of X[bar]= 0.77 was obtained.

If the app is correct, the true mean should be around cero, symbolically: μ=0

a. The hypotheses are:

H₀:μ=0

H₁:μ≠0

b. This test is a one-sample test for the population mean. To be able to do it you need the study variable to be at least normal. It is informed in the test that the population is normal, so the variable "difference between actual arrival time and estimated arrival time" has a normal distribution and the population variance is known, so you can conduct the test using the standard normal distribution.

c.

Z_{H_0}= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } }

Z_{H_0}= \frac{0.77-0}{\frac{2}{\sqrt{24} } }= 1.89

d. This hypothesis test is two-tailed and so is the p-value.

p-value: P(Z≤-1.89)+P(Z≥1.89)= P(Z≤-1.89)+(1 - P(Z≤1.89))= 0.029 + (1 - 0.971)= 0.058

e. 90% CI

Z_{1-\alpha /2}= Z_{0.95}= 1.645

X[bar] ± Z_{1-\alpha /2}* (\frac{Sigma}{\sqrt{n} } )

0.77 ± 1.645 * (\frac{2}{\sqrt{24} } )

[0.098;1.442]

I hope this helps!

4 0
3 years ago
Can someone solve n/7-12=5n+5 step by step pls
prohojiy [21]

Answer:

n=-7/2 I hate asking for things but can i have brainliest im trying to rank up.

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

n/7−12=5n+5

1/7n+−12=5n+5

1/7n−12=5n+5

Step 2: Subtract 5n from both sides.

1/7n−12−5n=5n+5−5n

-34/7n−12=5

Step 3: Add 12 to both sides.

-34/7n−12+12=5+12

-34/7n=17

Step 4: Multiply both sides by 7/(-34).

(7/-34)*(-34/7n)=(7/34)*(17)

3 0
3 years ago
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