The midpoint between x and 27 is -3. Find x.

For a filed trip the school bought 47 sandwiches for $40.60 each and 39 bags of chips for $1.25 each. How much did the school sp

lyudmila [28]

Answer:

$1,956.95

Step-by-step explanation:

The school bought 47 sandwiches for $40.60

= 47 × 40.60

= 1,908.2

They also bought 39 bags of chips for $1.25

= 1.25×39

= 48.75

Therefore the total amount spent can be calculated as follows

= 48.75 + 1,908.2

= 1,956.95

Hence the total amount spent by the school is $1,956.95

6 0

3 years ago

Can someone plz help me ASAP!

Zanzabum

Answer:

Step-by-step explanation:

Those bookmarks tho

7 0

3 years ago

Read 2 more answers

Please help me out here I would really appreciate it (will give brainiest if right)

Korvikt [17]

Answer:

265

Step-by-step explanation:

since take the 100 on the left if u put 2 lying down it wont fit, about half of that would fit. So it would be about 265

Mark as brainliest plz

4 0

3 years ago

4^x+2=1/2 solve for x (log equation)

lesantik [10]

$\log_ab=c\iff a^c=b\\\\4^{x+2}=\dfrac{1}{2}\qquad\text{use}\ a^{-1}=\dfrac{1}{a}\\\\(2^2)^{x+2}=2^{-1}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\2^{2(x+2)}=2^{-1}\iff2(x+2)=-1\qquad\text{use distributive property}\\\\(2)(x)+(2)(2)=-1\\\\2x+4=-1\qquad\text{subtract 4 from both sides}\\\\2x=-3\qquad\text{divide both sides by 2}\\\\\boxed{x=-1.5}$

3 0

3 years ago

Do one of the following, as appropriate. (a) Find the critical value z Subscript alpha divided by 2, (b) find the critical v

babymother [125]

Answer:

We want to construct a confidence interval at 99% of confidence, so then the significance level would be $1-0.99 =0.01$ and the value of $\alpha/2 =0.005$ . And for this case since we know the population deviation is not appropiate use the t distribution since we know the population deviation and the best quantile assuming that the population is normally distributed is given by the z distribution.

And if we find the critical value in the normal standard distribution or excel and we got:

$z_{\alpha/2}= \pm 2.576$

And we can use the following excel code:

"=NORM.INV(0.005,0,1)"

Step-by-step explanation:

For this case we have the following info given:

$\alpha=0.01, n =26, \sigma = 28.9$

We want to construct a confidence interval at 99% of confidence, so then the significance level would be $1-0.99 =0.01$ and the value of $\alpha/2 =0.005$ . And for this case since we know the population deviation is not appropiate use the t distribution since we know the population deviation and the best quantile assuming that the population is normally distributed is given by the z distribution.

And if we find the critical value in the normal standard distribution or excel and we got:

$z_{\alpha/2}= \pm 2.576$

And we can use the following excel code:

"=NORM.INV(0.005,0,1)"

7 0

3 years ago