List the partial products for this problem 252 x 605

tonya is twice kevin's age. in three years tonya will be 17. and in another three years, uncle rob will be three times tonyas ag

atroni [7]

Tonya is 14 (17 - 3)
<span>Kevin is 7 </span>
<span>Uncle Rob is 51 (3 x 17 = 54, less 3)</span>

3 0

3 years ago

Read 2 more answers

there are 400 pens in a shop 40percent are black 20percent of the other colored pens are blue and 10percent of other pens are ye

Over [174]

Percents is a number out of 100. So basically what you’re going to do here is add up the percents. 40+20+10. Add that up and you get 70. You then subtract 70 from 100. You get 30. So the percentage of red pens is 30%.

8 0

4 years ago

What type of angle is this?

GuDViN [60]

The correct answer is A which is obtuse

6 0

3 years ago

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.

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

The energy released by metabolism of 1 average candy bar is 1 x 10^6 joules.

Elis [28]

Answer: 32

Step-by-step explanation:

Given: The energy needed for 1 hour of running for an adult $=4.5\times10^6\ joules.$

Therefore. the energy needed for seven hours of running= $=7\times4.5\times10^6\ joules.=31.5\times10^6\ joules$

Since, he energy released by metabolism of 1 average candy bar $=1\times10^6\ joules.$

Therefore, the number of candy bars an adult need to eat to supply the energy needed for seven hours of running.= $\frac{31.5\times10^6}{1\times10^6}=31.5\approx32$

Hence, the adult need to eat 32 candies to supply the energy needed for seven hours of running.

3 0

3 years ago