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2 years ago

7

Members of the math club held a bake sale to raise money. Cupcakes and cookies were sold.

1 answer:

Mrac [35]2 years ago

4 0

Answer:

Cupcake

Step-by-step explanation:

Because the sale price is bigger

You might be interested in

If the measure of angle BCD=51 degrees, solve for x.

BabaBlast [244]

Answer:

$x=5$

Step-by-step explanation:

step 1

Find the measure of angle EFD

In this problem I will assume that ABCD is a parallelogram

In a parallelogram opposite angles are congruent and consecutive angles are supplementary

so

$m\ angle BCD=m\angle BED=51^o$

$m\ angle FED=(1/2)m\angle BED=(1/2)51^o=25.5^o$ --- > given problem

Remember that the sum of the interior angles in any triangle must be equal to 180 degrees

so

In the triangle EFD

$m\ angle FED+m\ angle FDE+m\ angle EFD=180^o$

substitute the given values

$25.5^o+55^o+m\ angle EFD=180^o$

$80.5^o+m\ angle EFD=180^o$

$m\ angle EFD=180^o-80.5^o$

$m\ angle EFD=99.5^o$

step 2

Find the measure of angle EFB

we know that

$m\angle EFB+m\angle EFD=180^o$ ---> by supplementary angles

we have

$m\ angle EFD=99.5^o$

substitute

$m\angle EFB+99.5^o=180^o$

$m\angle EFB=180^o-99.5^o$

$m\angle EFB=80.5^o$

step 3

Find the value of x

Remember that the sum of the interior angles in any triangle must be equal to 180 degrees

so

In the triangle EBF

$m\ angle BEF+m\ angle EFB+m\ angle EBF=180^o$

we have

$m\angle BEF=m\ angle FED=25.5^o$

$m\angle EFB=80.5^o$

$m\angle EBF=(14x+4)^o$

substitute

$25.5^o+80.5^o+(14x+4)^o=180^o$

solve for x

Combine like terms

$(14x+110)^o=180^o$

$14x=180-110$

$14x=70$

$x=5$

4 0

3 years ago

What is 3w(p +18) equal?

sashaice [31]

This problem calls for application of the distributive rule for multiplication.

<span>3w(p +18) can be expanded to 3wp + 54w.</span>

5 0

4 years ago

What is the answer to this (solve for a) 1a-15=4a-3

sleet_krkn [62]

Solve for a by simplifying both sides of the equation, then isolating the varible.
ANSWER: a= - 4
Hope this helps

4 0

4 years ago

A. y = 8 − 7x<br> B. y = 7x + 8<br> C. y = 7x − 8<br> D. y = 8x − 7

Anna71 [15]

Answer:

I'm getting

y=-7x+8

The 7 is negative for me..You should maybe try yourself too and see what you get I might have made a mistake

Step-by-step explanation:

I'm guessing you live somewhere in the U.S. so I'll use the equation you use there

Equation:

$y = mx + b$

$m = \frac{y1 - y2}{x1 - x2}$

first pick two points

I'll pick (0,8) and (1,1)

$m = \frac{8 - 1}{0 - 1} = \frac{7}{ - 1 } = - 7$

this means that the slope is -7

So far the equation will look like this

$y = - 7x + b$

to get b pick one point, I'll pick (0,8) and replace y and x

$8 = - 7 \times 0 + b$

when you solve this you'll get b

$b = 8$

So the equation will look this

$y = - 7x + 8$

7 0

3 years ago

A simple random sample of 100 8th graders at a large suburban middle school indicated that 81% of them are involved with some ty

Sophie [7]

Answer:

The interval is $0.7187 < p < 2.421$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 100$

The population proportion is $p = 0.81$

The confidence level is C = 98%

The level of significance is mathematically evaluated as

$\alpha = 100 -98$

$\alpha = 2%$ %

$\alpha = 0.02$

Here this level of significance represented the left and the right tail

The degree of freedom is evaluated as

$df = n-1$

substituting value

$df = 100 - 1$

$df = 99$

Since we require the critical value of one tail in order to evaluate the 98% confidence interval that estimates the proportion of them that are involved in an after school activity. we will divide the level of significance by 2

The critical value of $\frac{\alpha}{2}$ and the evaluated degree of freedom is

$t_{df , \alpha } = t_{99 , \frac{0.02}{2} } = 2.33$

this is obtained from the critical value table

The standard error is mathematically evaluated as

$SE = \sqrt{\frac{p(1-p )}{n} }$

substituting value

$SE = \sqrt{\frac{0.81(1-0.81 )}{100} }$

$SE = 0.0392$

The 98% confidence interval is evaluated as

$p - t_{df , \frac{\alpha }{2} } * SE < p < p + t_{df , \frac{\alpha }{2} }$

substituting value

$0.81 - 2.33 * 0.0392 < p < 0.81 +2.33 * 0.0392$

$0.7187 < p < 2.421$

4 0

3 years ago