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jenyasd209 [6]
3 years ago
10

Use the number line to represent -4 1/2 plus 3 1/4

Mathematics
1 answer:
Ilya [14]3 years ago
8 0

Start at - 4 1/2 adding a positive value moves to the right. Moving to the right 3 1/4 the slope becomes -1 1/4

Answer: -1 1/4

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What is the GCF of 4and 2
Hoochie [10]

Answer:

2

Step-by-step explanation:

GCF stands for the "greatest common factor." Therefore because 2 can go into both 2 and 4 evenly, it is the GCF.

3 0
3 years ago
Jackson flipped a coin ten time. He flipped heads 4 times and tails 6 times. What is the experimental probability that Jackson w
barxatty [35]

Answer:

Step-by-step explanation:

Given that,

Outcomes Heads= 4heads and

Outcomes Tails= 6tails

Total outcome of events n=10

Probability of head in is first throw is, the is the outcome of head divide by the total possible outcomes

P(Head in first throw)=4/10

P(H)=2/5

P(H) = 0.4

P(H) =40%

7 0
3 years ago
Read 2 more answers
An experimenter conducted a two-tailed hypothesis test on a set of data and obtained a p-value of 0.44. If the experimenter had
Lyrx [107]

Answer:

Correct option is (C).

The possible value of the <em>p</em>-value for a one-tailed test are 0.22 and 0.78.

Step-by-step explanation:

The <em>p</em>-value is the probability of acquiring a result as extreme as the observed result, assuming the null hypothesis statement is true.

The <em>p</em> value of a test is:

Left-tailed test: P(TS

Right-tailed test: P(TS>ts) = 1- P(TS.

Here,

TS = Test statistic

ts = computed value of the test statistic.

The two-tailed <em>p</em>-value is: 2P(TS or 2P(TS>ts).

The <em>p</em>-value of the two tailed test is, 0.44.

Compute the <em>p</em>-value for one-tailed test a follows:

  • For a left-tailed test:

            2P(TS

  • For a right-tailed test:

            P(TS>ts) = 1- P(TS

Thus, the possible value of the <em>p</em>-value for a one-tailed test are 0.22 and 0.78.

The correct option is (C).

3 0
3 years ago
Solve |2x + 2| /5 = -3
kaheart [24]

Answer:

there are no values of S

that make the equation true.

No solution

Step-by-step explanation:

5 0
3 years ago
Help, please! This is due tomorrow evening! Show work, please! Will give brainliest!
Maru [420]

Answer:

h = height reached

-16 t^2 = 1/2 g t^2 where g = 32 ft/sec^2

v t = height due to original vertical speed v

s = initial height

1. -45 = -16 t^2 + 62 t + 0         measuring height from original height

solving quadratic

2. t = 4.5 sec

3. time to reach max height = v / g = 62/32 = 1.94 sec

H (above release point) = -16 (1.94^2) + 62 * 1.94 = 60 ft

4. 2 * 1.94 = 3.88    solved in part 3

5. 1.94   solved on part 3

Check:  time to fall 60 + 45 ft = 105 ft

105 = 1/2 g t^2

t = (210 / 32)^1/2 = 2.56 sec

total time = 2.56 * 1.94 = 4.5 sec     time to fall + rise time

-45 = -16 * 4.5^2 + 62 * 4.5

-45 = -324 + 279 = -45        checking time to reach -45 feet using given equation

4 0
2 years ago
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