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11Alexandr11 [23.1K]

2 years ago

6

The teachers have $25 to buy supplies . Write and evaluate a numerical expression to show how much more money they will need to

buy the cups and napkins.
Cups= $3.50 -12 per package
Napkins= $4.25 - 10 per package

1 answer:

goldfiish [28.3K]2 years ago

3 0

I don't know man my teacher is gonna have to deal with me leaving this blank lol.

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What is the probability that a data value in a normal distribution is between a z-score of -1.32 and a z-score of -0.34 Round yo

Galina-37 [17]

We are asked to find the probability that a data value in a normal distribution is between a z-score of -1.32 and a z-score of -0.34.

The probability of a data score between two z-scores is given by formula $P(a$ .

Using above formula, we will get:

$P(-1.32$

Now we will use normal distribution table to find probability corresponding to both z-scores as:

$P(-1.32$

$P(-1.32$

Now we will convert $0.27351$ into percentage as:

$0.27351\times 100\%=27.351\%$

Upon rounding to nearest tenth of percent, we will get:

$27.351\%\approx 27.4\%$

Therefore, our required probability is 27.4% and option C is the correct choice.

5 0

3 years ago

How do you find the median and standard deviation for 3,3,3,3,4,4,4,4,4,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6,7,7,7,7,7,7,8,8,8,8,8,

Rashid [163]

Answer:

Median is 6

standard deviation is 2.39

Step-by-step explanation:

the median is the mid number in the data set

The median formula is {(n + 1) ÷ 2}th, where “n” is the number of items in the set and “th” just means the (n)th number.

There are 51 data points in this set (n = 51)

so 51 / 2 = 25.5, the. number in the 25th slot is our median

3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6,<u> 6</u>, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11

the standard deviation is a measure of how dispersed the data is in relation to the mean. Low standard deviation means data are clustered around the mean, and high standard deviation indicates data are more spread out.

7 0

2 years ago

Solve the following system of equations for all three variables.

xz_007 [3.2K]

Answer:

Step-by-step explanation:

-2x + 3y – 4z = 8 ----------------(I)

5x – 3y + 5z = -8 -----------------(II)

7x – 3y + 3z = 8 ------------------(III)

Add equation (I) & (II) and thus y will be eliminated

(I) -2x + 3y – 4z = 8

(II) <u>5x – 3y + 5z = -8</u> {Add}

3x + z = 0 ------------------------(A)

Multiply equation (II) by (-1) and then add with equation (III). Thus y will be eliminated.

(II) * (-1) -5x + 3y - 5z = +8

<u>7x – 3y + 3z = 8</u> {Add}

2x -2z = 16 ---------------(B)

Multiply equation (A) by 2 and then add. Thus z will be eliminated and we will get the value of x

(A) * 2 6x + 2z = 0

(B) <u>2x - 2z = 16</u> {Add}

8x = 16

Divide both sides by 8

x = 16/8

x = 2

Plugin x = 2 in equation (A)

3x + z = 0

3*2 + z = 0

6 + z = 0

z = -6

Plug in x = 2 and z = - 6 in equation (I)

-2x +3y - 4z = 8

-2*2 + 3y - 4*(-6) = 8

-4 + 3y + 24 = 8

3y + 20 = 8

3y = 8 - 20

3y = -12

y = -12/3

y = -4

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7 0

3 years ago

Read 2 more answers

Find the derivative.

Aleksandr [31]

Answer:

Using either method, we obtain: $t^\frac{3}{8}$

Step-by-step explanation:

a) By evaluating the integral:

$\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du$

The integral itself can be evaluated by writing the root and exponent of the variable u as: $\sqrt[8]{u^3} =u^{\frac{3}{8}$

Then, an antiderivative of this is: $\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}$

which evaluated between the limits of integration gives:

$\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}$

and now the derivative of this expression with respect to "t" is:

$\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}$

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

$g(x)=\int\limits^x_a {f(t)} \, dt$

is continuous on [a,b], differentiable on (a,b) and $g'(x)=f(x)$

Since this this function $u^{\frac{3}{8}$ is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

$\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}$

5 0

3 years ago

1. Find the capacity in litres of a bucket 24cm in

Sav [38]

Answer:

6.36l

Step-by-step explanation:

We compute first the volume:

V=pi×h/3×(R²+r²+rR), where h=20cm is the height, R=24/2=12cm is the top radius, r=16/2=8cm is the bottom radius.

We get: V=pi×20/3(12²+8²+12×8)=

20×pi/3(144+64+96)cm³

V=20×pi/3×304=6363.73cm³

In dm³ we have (divide by 1000) V=6.36dm³

By definition, 1dm³=1l, so the capacity is 6.36l

8 0

3 years ago