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TEA [102]
4 years ago
7

the line plot shows the number of different colors that art students used in a colored pencil sketch. how many art students used

more than 6 colors ​

Mathematics
2 answers:
Tomtit [17]4 years ago
5 0

Answer:

The number of art students used more than 6 colors = 8.

Step-by-step explanation:

The line plot shows the number of different colors that art students used in a colored pencil sketch.

The number of students used 7 colors = 1

The number of students used 8 colors = 2

The number of students used 9 colors = 4

The number of students used 10 colors = 1

We need to find the sum = 1 + 2 + 4 + 1 = 8

So the number of art students used more than 6 colors = 8.

stiks02 [169]4 years ago
4 0

Answer:

8

Step-by-step explanation:

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3 years ago
Randy, Sandy and Todd are my cousins and their combined age is 87 years. Randy is four times as old as Sandy, but Todd is actual
NISA [10]

Answer:

Randy is 48 years old

Sandy is 12 years old

Todd is 27 years old

<em></em>

Step-by-step explanation:

Given

Represent their ages with the first letters of their names:

R + S + T = 87

R = 4 * S

T = 15 + S

Required

Determine their individual ages

We'll solve the above equations using substitution method of simultaneous equations

Substitute 15 + S for T and 4 * S for R in R + S + T = 87

This gives:

4 * S + S + 15 + S = 87

4 S + S + 15 + S = 87

Collect Like Terms

4 S + S + S = 87 - 15

6S = 72

Divide through by 6

S = 72/6

S = 12

<em>This implies that Sandy is 12 years old</em>

Substitute 12 for S in R = 4 * S and  T = 15 + S

R = 4 * 12

R = 48

T = 15 + 12

T = 27

<em>This implies that Randy is 48 years old and Todd is 27</em>

7 0
3 years ago
Emily sold 148 boxes of cookies for a fundraiser. If she sold 80% of the total number of boxes sold for the fundraiser, how many
IRINA_888 [86]

Answer:

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7 0
3 years ago
Read 2 more answers
A university is interested in whether there's a difference between students who live on
ryzh [129]

It is to be noted that the determination of whether or not there is a significant difference between the two groups will be done using a t test.

<h3>What is a t test?</h3>

A t-test is a statistical test that juxtaposes two samples' means. It is used in hypothesis testing, using a null hypothesis that the variance in group means is zero and an alternative hypothesis that the difference is not zero.

<h3>What are the conditions for using this kind of confidence interval? </h3>

The conditions to use the t test are:

  • The sample must be independent
  • The mean of the population and variance must be unknown.
  • The Box plot is attached.

<h3>What are the degrees of freedom (k) for this test using the conservative method?</h3>

The degrees of freedom (k) to be utilized for this text will be derived using the conservative method given below:

df = [(s₁²/n₁) + (s₂²/n²)/[((s₁²/n₁)²/((n₁-1)) + (s₂²/n₂)²/((n₂-1))]

= [(3.0952/15) + (6.4095/15)]² / [((3.0952/15)²/14) + ((6.4095/15)²/14)]

= 24.965

Hence,

df ≈ 24 (if approximated to the floor)

<h3>What are the sample statistics for this test?</h3>

Recall the the standard deviation of the population are unequal and unknown. This thus requires that we utilize the two-sample unpooled t-test.

Here, H₀ is given as;

t = \frac{\bar{x_{1} -\bar{x_{2}}}}{\sqrt{\frac{s_{1}^{2} }{n_{1} } + \frac{S_{2}^{2} }{n_{2}}  } } \sim  t_{df}

t = [(3.33333 - 4.13333)]/√[(3.0952/15) + (6.4095/15)]

= - 0.8/√0.6337

t = - 1.005

<h3>What is the 95% confidence interval for the difference between the number of classes missed by each group of students?</h3>

The 95% confidence interval is computed using the following formula:

(\bar x_{2} - \bar x_{1}) \pm t_\alpha_/_2_,_df \left({\sqrt{\frac{S_{1}^{2} }{n_{1} } + \frac{S_{2}^{2} }{n_{2}}  } } \right)

= - 0.8 ± t₀.₀₂₅,₂₄ (√0.6337)

= - 0.8 ± 2.064 (√0.6337)

= -2.4429, 0.8429

<h3>What is the a 90% confidence interval for the difference between the number of classes missed by each group of students?</h3>

To derive the 90% interval, we state:

(\bar x_{2} - \bar x_{1}) \pm t_\alpha_/_2_,_df \left({\sqrt{\frac{S_{1}^{2} }{n_{1} } + \frac{S_{2}^{2} }{n_{2}}  } } \right)

= - 0.8 ± t₀.₀₅₀,₂₄ (√0.6337)

= - 0.8 ± 0.685 (√0.6337)

= -2.162, 0.562

<h3>Based on the two confidence intervals computed in parts d and e, what is the conclusion about the differences between the means of the two groups?</h3>

From the intervals computed, we must fail to reject H₀

H₀ : μ₁ = μ₂

It is clear from the above intervals computed from that the differences between the mean of both groups is significant. This is because, zero is included on the two intervals.

Learn more about t-test at;
brainly.com/question/6589776
#SPJ1

8 0
2 years ago
Solve the problems.
alekssr [168]
C...........................
5 0
3 years ago
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