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Crank
3 years ago
5

Multipying these is a pain for me. Please help.

Mathematics
1 answer:
stiks02 [169]3 years ago
5 0

Answer:

a10 b4

Step-by-step explanation:

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7. On Thursday, Kim purchased a medium iced caramel latte and a blueberry muffin for $5.18. Then, on Friday, Kim
sashaice [31]

Answer:

Cost of medium iced caramel latte = $3.59

Step-by-step explanation:

Assume;

Cost of medium iced caramel latte = x

Cost of blueberry muffin = y

So,

On Thursday

x + y = 5.18......eq1

On Friday

2x + 3y = 11.95.....eq2

Eq1 × 3

3x + 3y = 15.54......eq3

eq3 - eq2

x = 3.59

Cost of medium iced caramel latte = $3.59

4 0
3 years ago
How do you do the area and how to do it to the nearest tenth
lord [1]
You just put I line in the middle and you times high times base the do the other one then add your answers  up that should be your finely answer 
6 0
3 years ago
Is a + b rational or irrational
Crank

Answer and fdaStep-by-step explanation:::

The answer depends on the numbers that are inputted for a and b.

If the numbers that are inputted for both a and b are rational, then a + b would be rational.

If the numbers that are inputted for both a and b are irrational, then a + b would be irrational.

If one of the variables is rational and the other is irrational, then a + b would be irrational.

#teamtrees #PAW (Plant And Water)

6 0
3 years ago
The following data lists the ages of a random selection of actresses when they won an award in the category of Best​ Actress, al
Valentin [98]

Answer:

a) p_v =P(t_{(9)}

The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best​ Actors at 1% of significance.

b) The 99% confidence interval would be given by (-21.469;2.069)

c) We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two

Step-by-step explanation:

Part a

Let put some notation  

x=actor's age , y = actress's age

x: 58 41 36 36 34 33 48 37 37 43

y: 26 27 34 26 35 29 23 42 30 34

The system of hypothesis for this case are:

Null hypothesis: \mu_y- \mu_x \geq 0

Alternative hypothesis: \mu_y -\mu_x

The first step is calculate the difference d_i=y_i-x_i and we obtain this:

d: -32, -14, -2, -10, 1, -4, -25, 5, -7, -9

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= -9.7

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =11.451

The 4 step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-9.7 -0}{\frac{11.451}{\sqrt{10}}}=-2.679

The next step is calculate the degrees of freedom given by:

df=n-1=10-1=9

Now we can calculate the p value, since we have a left tailed test the p value is given by:

p_v =P(t_{(9)}

The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best​ Actors at 1% of significance.

Part b

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

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

\bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,9)".And we see that t_{\alpha/2}=3.25  

Now we have everything in order to replace into formula (1):  

-9.7-3.25\frac{11.451}{\sqrt{10}}=-21.469  

-9.7+3.25\frac{11.451}{\sqrt{10}}=2.069  

So on this case the 99% confidence interval would be given by (-21.469;2.069)

Part c

We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two means is 0.

8 0
3 years ago
A random sample of 100 automobile owners in thestate of Virginia shows that an automobile is driven onaverage 23,500 kilometers
Anastaziya [24]

Answer:

a) 22497.7 < μ< 24502.3

b)  With 99% confidence the possible error will not exceed 1002.3

Step-by-step explanation:

Given that:

Mean (μ) = 23500 kilometers per year

Standard deviation (σ) = 3900 kilometers

Confidence level (c) = 99% = 0.99

number of samples (n) = 100

a) α = 1 - c = 1 - 0.99 = 0.01

\frac{\alpha }{2} =\frac{0.01}{2}=0.005\\ z_{\frac{\alpha }{2}}=z_{0.005}=2.57

Using normal distribution table, z_{0.005 is the z value of 1 - 0.005 = 0.995 of the area to the right which is 2.57.

The margin of error (e) is given as:

e= z_{0.005}\frac{\sigma}{\sqrt{n} }  = 2.57*\frac{3900}{\sqrt{100} } =1002.3

The 99% confidence interval = (μ - e, μ + e) = (23500 - 1002.3, 23500 + 1002.3) =  (22497.7, 24502.3)

Confidence interval = 22497.7 < μ< 24502.3

b) With 99% confidence the possible error will not exceed 1002.3

5 0
3 years ago
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