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Nuetrik [128]
3 years ago
15

PLEASE HELP ASAP! 50 POINTS!! HELP HELP HELP!!

Mathematics
1 answer:
marin [14]3 years ago
7 0
\frac{ \frac{8}{5} + ( - \frac{5}{6}) }{ \frac{1}{6} }  \\ = \frac{ \frac{8}{5} - \frac{5}{6} }{ \frac{1}{6} } \\ = \frac{ \frac{48- 25}{30} }{ \frac{1}{6} } \\ = \frac{ \frac{23}{30} }{ \frac{1}{6} } \\ = \frac{23}{30} \times 6 \\ = \frac{23}{5}
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Suppose that the water level of a river is 34 feet and that it is receding at a rate of
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The answer is C I just took the test
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Describe how k affects the graph of the function f(x)= -3/5(10)^x
Shalnov [3]

Answer:

1437.33

Step-by-step explanation:

8 0
3 years ago
A peony plant with red petals was crossed with another plant having streaky petals. A geneticist states that 75% of the offsprin
Strike441 [17]

Answer:

The sample data confirm the geneticist's prediction that 75% of the offspring from this cross will have red flowers.

Step-by-step explanation:

A Chi-square goodness of fit test can be used to test the claim made by the geneticist.

The hypothesis is defined as:

<em>H₀</em>: There is no difference between the observed and expected value,i.e. <em>p</em>₁ < 0.75.

<em>Hₐ</em>: There is a significant difference between the observed and expected value, i.e. <em>p</em>₁ > 0.75.

The test statistic is:

\chi^{2}=\sum \frac{(O-E)^{2}}{E}

Consider the tables attached below.

The value of the test statistic is:

\chi^{2}=\sum \frac{(O-E)^{2}}{E}=25.8133

The critical value is:

\chi^{2}_{\alpha, k-1}=\chi^{2}_{0.01, 1}=6.635

*Use a Chi-square table.

The critical region is:

\chi^{2}\geq 6.635

The test statistic value of 25.8133 lies in the critical region.

The null hypothesis is rejected at 1% level of significance.

<u>Conclusion</u>:

As the null hypothesis is rejected it can be concluded that 75% of the offspring from this cross will have red flowers.

7 0
3 years ago
1.
Vesnalui [34]
X is 125 as well because it’s an Cubs all sides are equal
4 0
3 years ago
Use technology or a z-score table to answer the question.
Levart [38]

Answer: A:60%

Step-by-step explanation:

Since the scores for a golf tournament are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = scores for the tournament.

µ = mean score

σ = standard deviation

From the information given,

µ = 210

σ = 80

We want to find the probability percent of golfers that scored less than Ella. It is expressed as

P(x < 230)

z = (230 - 210)/80 = 0.25

Looking at the normal distribution table, the probability corresponding to the z score is 0.5987

Therefore, the percent of golfers that scored less than Ella is

0.5987 × 100 = 59.87

Approximately 60%

6 0
4 years ago
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