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Nikolay [14]
3 years ago
8

Katie needs to buy food for her pet leopard gecko the cost of the food is $3.79 if katie uses a coupon the costs is reduced to $

3.03 how much lower is the cost with the coupon than without it in cents
Mathematics
1 answer:
LekaFEV [45]3 years ago
8 0
The cost with the coupon is $0.76 less.

3.79 - 3.03
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(Help plz)
Aneli [31]

Answer:

86.7 ft²

Step-by-step explanation:

arc area = area of the circle x ( arc angle measure / 360°)

area of the circle = 65 / (270/360) = 86.67 ft²

3 0
3 years ago
Solve the equation: -2x(x+8)+24=11-3x
Murrr4er [49]
X=4 I think I’m not sure
8 0
3 years ago
Read 2 more answers
What expression is equal to 6×8?<br> A. 8×6<br><br> B. 14×1<br><br> C. 6×4×8<br><br> D. 4+2×7+1
iogann1982 [59]
Well, you need to see which is equal to 6*8
6*8 = 48

14*1 = 14
6*4*8 = 192
(4+2)*(7+1) = 48
6 0
3 years ago
2. what value of b makes the following equation true
Svetradugi [14.3K]

Answer:

  2. 10

  3. D.  1 for all n

Step-by-step explanation:

2. The applicable rules of exponents are ...

  (a^b)(a^c) = a^(b+c)

  a^b = 1/a^-b

__

  \dfrac{4^8}{(4^2)^{-3}}\div 4^4=\dfrac{4^8}{4^{-6}\cdot4^4}=\dfrac{4^8}{4^{-2}}\\\\=4^8\cdot4^2=4^{10}

The value of n is 10.

__

3. Using the above rules of exponents, the expression simplifies to ...

  6^(-n+n) = 6^0 = 1

The value is 1 for any n.

4 0
3 years ago
A manager is comparing wait times for customers in a coffee shop based on which employee is
anyanavicka [17]

Using the t-distribution, as we have the standard deviation for the sample, it is found that there is a significant difference between the wait times for the two populations.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, we test if there is no difference, that is:

H_0: \mu_A - \mu_B = 0

At the alternative hypothesis, it is tested if there is difference, that is:

H_1: \mu_A - \mu_B = 0

<h3>What are the mean and the standard error of the distribution of differences?</h3>

For each sample, we have that:

\mu_A = 73, s_A = \frac{2}{\sqrt{100}} = 0.2

\mu_B = 74, s_B = \frac{4}{\sqrt{100}} = 0.4

For the distribution of differences, we have that:

\overline{x} = \mu_A - \mu_B = 73 - 74 = -1

s = \sqrt{s_A^2 + s_B^2} = \sqrt{0.2^2 + 0.4^2} = 0.447

<h3>What is the test statistic?</h3>

It is given by:

t = \frac{\overline{x} - \mu}{s}

In which \mu = 0 is the value tested at the null hypothesis.

Hence:

t = \frac{\overline{x} - \mu}{s}

t = \frac{-1 - 0}{0.447}

t = -2.24

<h3>What is the p-value and the decision?</h3>

Considering a one-tailed test, as stated in the exercise, with 100 - 1 = 99 df, using a t-distribution calculator, the p-value is of 0.014.

Since the p-value is less than the significance level of 0.05, it is found that there is a significant difference between the wait times for the two populations.

More can be learned about the t-distribution at brainly.com/question/16313918

8 0
2 years ago
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