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kompoz [17]
3 years ago
7

THIS IS THE THRID TIMEE PLZZZ A card was selected at random from a standard deck of cards. The suit of the card was recorded, an

d then the card was put back in the deck. The table shows the results after 40 trials.What is the relative frequency of selecting a heart? 15% 25% 27% 35% outcome 8 12 14 6
Mathematics
1 answer:
Oxana [17]3 years ago
4 0

Answer:

The answer to your question is C (27%)

Step-by-step explanation:

I looked up the answer for you but this was proven to be correct.

You might be interested in
D divided bye 1.4=2.3 what is d
kati45 [8]
Multiply 2.3 by 1.4= D
4 0
3 years ago
Read 2 more answers
the second of two number is 6 more than 3 times the first. The sum of the two number is 30. Find the two numbers​
pantera1 [17]

Answer:

kinda hard

Step-by-step explanation:

7 0
3 years ago
A total of h males and (h - 2) females visit the museum on a certain day. If total of visitors for that day is 80, what is the v
Akimi4 [234]

Given:

Number of males = h

Number of females = (h-2)

Total number of visitors = 80

To find:

The value of h.

Solution:

We have,

Number of males = h

Number of females = (h-2)

So, the total number of visitors is:

Total=h+(h-2)

Total=2h-2

It is given that the total number of visitors is 80. So,

2h-2=80

2h=80+2

2h=82

Divide both sides by 2.

h=\dfrac{82}{2}

h=41

Therefore, the correct option is D.

3 0
3 years ago
Find the p-value: An independent random sample is selected from an approximately normal population with an unknown standard devi
vladimir1956 [14]

Answer:

(a) <em>p</em>-value = 0.043. Null hypothesis is rejected.

(b) <em>p</em>-value = 0.001. Null hypothesis is rejected.

(c) <em>p</em>-value = 0.444. Null hypothesis is not rejected.

(d) <em>p</em>-value = 0.022. Null hypothesis is rejected.

Step-by-step explanation:

To test for the significance of the population mean from a Normal population with unknown population standard deviation a <em>t</em>-test for single mean is used.

The significance level for the test is <em>α</em> = 0.05.

The decision rule is:

If the <em>p - </em>value is less than the significance level then the null hypothesis will be rejected. And if the <em>p</em>-value is more than the value of <em>α</em> then the null hypothesis will not be rejected.

(a)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> > <em>μ₀</em>

The sample size is, <em>n</em> = 11.

The test statistic value is, <em>t</em> = 1.91 ≈ 1.90.

The degrees of freedom is, (<em>n</em> - 1) = 11 - 1 = 10.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 1.90 and degrees of freedom 10 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₁₀ > 1.91) = 0.043.

The <em>p</em>-value = 0.043 < <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

(b)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> < <em>μ₀</em>

The sample size is, <em>n</em> = 17.

The test statistic value is, <em>t</em> = -3.45 ≈ 3.50.

The degrees of freedom is, (<em>n</em> - 1) = 17 - 1 = 16.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of -3.50 and degrees of freedom 16 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₁₆ < -3.50) = P (t₁₆ > 3.50) = 0.001.

The <em>p</em>-value = 0.001 < <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

(c)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> ≠ <em>μ₀</em>

The sample size is, <em>n</em> = 7.

The test statistic value is, <em>t</em> = 0.83 ≈ 0.82.

The degrees of freedom is, (<em>n</em> - 1) = 7 - 1 = 6.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 0.82 and degrees of freedom 6 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₆ < -0.82) + P (t₆ > 0.82) = 2 P (t₆ > 0.82) = 0.444.

The <em>p</em>-value = 0.444 > <em>α</em> = 0.05.

The null hypothesis is not rejected at 5% level of significance.

(d)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> > <em>μ₀</em>

The sample size is, <em>n</em> = 28.

The test statistic value is, <em>t</em> = 2.13 ≈ 2.12.

The degrees of freedom is, (<em>n</em> - 1) = 28 - 1 = 27.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 0.82 and degrees of freedom 6 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₂₇ > 2.12) = 0.022.

The <em>p</em>-value = 0.444 > <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

5 0
3 years ago
Perform the following computations.You may use13≈0.333333,34= 0.75 and100301= 0.332226.(i). Compute13+34by using five significan
White raven [17]

Answer:

a. 1.0833

Absolute Error = 0.416667

Relative Error = 1.250002

b. 0.0011070

Absolute Error = 0.0011070

Relative Error = 0.003321

Step-by-step explanation:

Given

1/3 = 0.333333

3/4 = 0.75

100/301 = 0.332226

a.

1/3 + 3/4

= 0.333333 + 0.75

= 1.083333

= 1.0833 ------ Approximated to 5 significant digits

Absolute Error = |Real Value - Estimated Value|

Relative Error = Absolute Error/Real Value

Assume 1/3 to be the real value and 3/4 to be the estimated value

Absolute Error = |0.333333 - 0.75|

Absolute Error = |-0.416667|

Absolute Error = 0.416667

Relative Error = 0.416667/0.333333

Relative Error = 1.250002

b.

1/3 - 100/301

= 0.333333 - 0.332226

= 0.001107

= 0.0011070 ----- Approximated to 5 significant digits

Assume 1/3 to be real value and 100/301 to be estimated value

Absolute Error = 0.333333 - 0.332226

Absolute Error = 0.0011070

Relative Error = 0.0011070/0.333333

Relative Error = 0.003321

Absolute and relative errors are approximation errors and they are due to the discrepancy between an exact value and some approximation to them.

The absolute error is the magnitude of the difference between the exact value and the approximation. The relative error is the absolute error divided by the magnitude of the exact value

4 0
3 years ago
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