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GrogVix [38]
3 years ago
11

Pls help me I really need help will give brainlest

Mathematics
2 answers:
Lynna [10]3 years ago
8 0

She pays a total of $615 in taxes each month and her monthly net income is 1,065 a month

ValentinkaMS [17]3 years ago
3 0

420+86+109=615 I hope this helps you

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What is the surface area of a cube that has a side length of 5 feet?
harkovskaia [24]

Hello!

<h3><em><u>Answer</u></em></h3><h3 />

The surface area of the cube is 150 ft^{2}.


<h3><em><u>Explanation</u></em></h3>

A = 6 a^{2}

A = 6 5^{2}

A = 6 × 25

A = 150

6 0
3 years ago
Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):
Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, \mu_X represent the population mean for Brand B and let \mu_Y represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

        P.Q. = \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } ~ t_n__1+n_2-2

where, Xbar = Sample mean for Brand B data = 36.9

            Ybar = Sample mean for Brand A data = 32.9

              n_1  = Sample size for Brand B data = 13

              n_2 = Sample size for Brand A data = 9

              s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2}  }{n_1+n_2-2} } = \sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} } = 3.013

Here, s^{2}_X and s^{2} _Y are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < t_2_0 < 2.528) = 0.98

P(-2.528 < \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } < 2.528) = 0.98

P(-2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (Xbar -Ybar) -(\mu_X-\mu_Y) < 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

P( (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (\mu_X-\mu_Y) < (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} , (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ]

[ (36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} , (36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

                     

4 0
3 years ago
Three numbers have an average of 15. If two of the numbers are 10 and 12, then what is the other number?
AveGali [126]

Answer:

Step-by-step explanation:

(12+10+x)/3=15

15*3=12+10+x

45-12-20=x

X=23

3 0
2 years ago
Multiplying monomials and binomials
lubasha [3.4K]

Answer:

28w^2-476w

Step-by-step explanation:

The general rule we are going to use to multiply this out is the distributive property. Which is:

a(b+c) = ab + ac

<u>Note:</u> x * x = x^2

Now multiplying, we get:

28w(w-17)\\=28w*w-28w*17\\=28w^2-476w

This is the multiplied out form, answer.

7 0
3 years ago
I chose circle is closed and graph is shaded to the left of the circle and it said that was wrong. Please help, thank you
Tems11 [23]
The circle is closed and the graph is shaded to the right also there’s a negative on the x, you have to get rid of it first by dividing by -1 and then you switch direction of the less than or equal to greater than or equal because you divided by negative

6 0
3 years ago
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