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tensa zangetsu [6.8K]
2 years ago
13

When we first learn arithmetic, we focus on working with whole numbers. Then, we extend our understanding to fractions and negat

ive numbers. In algebra, we transfer these same skills to expressions involving variables. What elementary skills have you used so far in this course, and how have you extended them?
Mathematics
1 answer:
Alexeev081 [22]2 years ago
5 0

Answer:

Well, I don't think I can remember EVERY skill I was taught through the years but in Kindergarten we used Number Sense so we could accurately... also there's estimation and of course PEMDAS. I've also learned how to round up so that the solution is at least around those numbers if the skill is to be applied.

Step-by-step explanation:

I'm not sure if this'll help but go ahead and have it... just try and change it a bit so they don't get ya for plagirism

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Eighty percent of one store's customers paid with credit cards. Forty customers came in that day. How many customers paid for th
mario62 [17]
80%=.8

40*.8=32

32 people made their purchases with credit cards.
6 0
2 years ago
Read 2 more answers
In a West Texas school district the school year began on August 1 and lasted until May 31. On August 1 a Soft Drink company inst
WINSTONCH [101]

Answer:

$95.78

Step-by-step explanation:

f(t) = 300t / (2t² + 8)

t = 0 corresponds to the beginning of August.  t = 1 corresponds to the end of August.  t = 2 corresponds to the end of September.  So on and so forth.  So the second semester is from t = 5 to t = 10.

$T₂ = ∫₅¹⁰ 300t / (2t² + 8) dt

$T₂ = ∫₅¹⁰ 150t / (t² + 4) dt

$T₂ = 75 ∫₅¹⁰ 2t / (t² + 4) dt

$T₂ = 75 ln(t² + 4) |₅¹⁰

$T₂ = 75 ln(104) − 75 ln(29)

$T₂ ≈ 95.78

3 0
2 years ago
A manufacturing company produces valves in various sizes and shapes. One particular valve plate is supposed to have a tensile st
maxonik [38]

Answer:

z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413    

p_v =2*P(z>1.413)=0.158  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

Step-by-step explanation:

Data given and notation  

\bar X=5.0611 represent the sample mean

\sigma=0.2803 represent the population standard deviation for the sample  

n=42 sample size  

\mu_o =5 represent the value that we want to test

\alpha=0.1 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 5, the system of hypothesis would be:  

Null hypothesis:\mu = 5  

Alternative hypothesis:\mu \neq 5  

If we analyze the size for the sample is > 30 but and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413    

P-value

Since is a two sided test the p value would be:  

p_v =2*P(z>1.413)=0.158  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

8 0
2 years ago
Pecans that cost $28.50 per kilogram were mixed with almonds that cost $22.25 per kilogram. How many kilograms of each were used
atroni [7]

Answer:

The pecans used in the mixture =  8 kg

The almonds used in the mixture  =  17 kg

Step-by-step explanation:

The cost of per kg of pecans  = $28.50

The cost of per kg of almonds = $22.25

Now, total weight of mixture  = 25 kg

Let us assume the weight of pecans in the mixture   =   m kg

So, the amount of almonds in the mixture  = Total weight - Weight of Pecans

= (25 - m) kg

Now, cost of m kg pecans  = m x ( cost of 1 kg pecans) = m x ( $ 28.50)

                                              =  28.50 m

cost of (25- m) kg almonds  = (25 -m)  x ( cost of 1 kg almonds)

                                                = (25 - m) x ( $ 22.25)

                                                =  556.25  - 22.25 m

Total cost  of 25 kg of mixture  = 25 x ( $24.25) =  $606.25

Now, Total Cost of (almonds  + Pecans)  = Total cost of mixture

⇒28.50 m + 556.25  - 22.25 m  = $606.25

or,6.25 m =  50

or, m = 50/6.25 =  8

Hence, the pecans used in the mixture = m = 8 kg

And the  almonds used in the mixture  = (25 - m) = 25 - 8 = 17 kg

5 0
2 years ago
6.In a sample of 131 women with cosmetic dermatitis from using eye shadow, 12 were diagnosed with a nickel allergy. In a sample
aivan3 [116]

Answer:

a) 0.0916 - 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.0422

0.0916 + 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.1410

So then we can conclude that the true proportion of women with cosmetic dermatitis from using eye shadow at 95% of confidence is between (0.0422 and 0.1410)

b) \hat p = \frac{25}{250}= 0.1

0.1 - 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.0628

0.1 + 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.1372

So then we can conclude that the true proportion of women with cosmetic dermatitis from using mascara at 95% of confidence is between (0.0628 and 0.1372)

c) For this case we see that both confidence intervals contains the value of 0.12 so then we can't conclude that only one group is referenced at the significance level of 0.05 used.

Step-by-step explanation:

Part a

The estimated proportion of women with cosmetic dermatitis from using eye shadow is given by:

\hat p =\frac{12}{131}= 0.0916

The confidence  interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The confidence interval for the proportion is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

Replacing we got:

0.0916 - 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.0422

0.0916 + 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.1410

So then we can conclude that the true proportion of women with cosmetic dermatitis from using eye shadow at 95% of confidence is between (0.0422 and 0.1410)

Part b: A 95% confidence interval for the women with cosmetic dermatitis from using mascara

\hat p = \frac{25}{250}= 0.1

0.1 - 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.0628

0.1 + 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.1372

So then we can conclude that the true proportion of women with cosmetic dermatitis from using mascara at 95% of confidence is between (0.0628 and 0.1372)

Part c: Suppose you are informed that the true proportion with a nickel allergy for one of the two groups (eye shadow or mascara) is .12. Can you determine which group is referenced? Explain.

For this case we see that both confidence intervals contains the value of 0.12 so then we can't conclude that only one group is referenced at the significance level of 0.05 used.

7 0
2 years ago
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