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zhannawk [14.2K]
2 years ago
10

if a bus holds 40 students. if the school fills 3 1/2 buses for this field trip how many students are there

Mathematics
1 answer:
Alborosie2 years ago
7 0

Answer:

140 students......

Step-by-step explanation:

Plz mark brainliest thanks

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Find x step by step (10 points)
AleksandrR [38]

Answer:

\sqrt{(n-m)(n+m)} & -\sqrt{(n-m)(n+m)}

Step-by-step explanation:

1) Subtract m^{2} from both sides. This should leave you with x^{2}=n^{2}-m^{2}.

2) Square root both sides. This should leave you with x=\sqrt{n^2-m^2} & x=-\sqrt{n^2-m^2}.

<em>You can stop here if this is what the problem is asking for. However, it is not fully simplified.</em>

<em />

3) Factor the equation. This should leave you with \sqrt{(n-m)(n+m)} & -\sqrt{(n-m)(n+m)}.

4 0
2 years ago
Read 2 more answers
Harold has been hired as a salesman. He is paid a flat rate of $375 each week and earns an additional $14.25 commission for each
Andru [333]
The answer is
c. \: y = 14.25h + 375
5 0
3 years ago
A researcher is interested in determining if the more than two thirds of students would support making the Tuesday before Thanks
Yakvenalex [24]

Answer:

The null hypothesis is, {H₀}: p = 0.66, H₀: p = 0.66 .

The alternative hypothesis is, {Hₐ}: p > 0.66, Hₐ​: p > 0.66

Step-by-step explanation:

Let p represent the proportion of students who are supportive of making the day before Thanksgiving a holiday.

Since two populations are considered, the alternative hypothesis for the difference of the means of the two populations has to be stated.

The proportion of two-third students are supportive of making the day before Thanksgiving a holiday is p = 0.66 obtained as shown below:

The null hypothesis is,

{H₀}:p = 0.66, H₀​: p = 0.66

The alternative hypothesis is,

{Hₐ:p > 0.66, Ha​: p > 0.66

The null hypothesis is, {H₀}: p = 0.66, H₀: p=0.66 .

The alternative hypothesis is, {Hₐ}: p > 0.66, Hₐ​: p > 0.66

3 0
3 years ago
The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.0 minutes and
strojnjashka [21]

Answer:

P ( 5 < X < 10 ) = 1

Step-by-step explanation:

Given:-

- Sample size n = 49

- The sample mean u = 8.0 mins

- The sample standard deviation s = 1.3 mins

Find:-

Find the probability that the average time waiting in line for these customers is between 5 and 10 minutes.

Solution:-

- We will assume that the random variable follows a normal distribution with, then its given that the sample also exhibits normality. The population distribution can be expressed as:

                                   X ~ N ( u , s /√n )

Where

                            s /√n = 1.3 / √49 = 0.2143

- The required probability is P ( 5 < X < 10 ) minutes. The standardized values are:

                        P ( 5 < X < 10 ) = P (    (5 - 8) / 0.2143 <  Z  <  (10-8) / 0.2143   )

                                                 = P ( -14.93 < Z < 8.4 )

- Using standard Z-table we have:

                        P ( 5 < X < 10 ) = P ( -14.93 < Z < 8.4 ) = 1        

7 0
3 years ago
A tin of biscuits contains shortbread and choc-chip biscuits in the ratio 2:3 . The tin contains 80 biscuits. How many short-bre
Aleksandr [31]

Answer:

32

Step-by-step explanation:

There are 80 biscuits inside the tin.

Also, they said that those biscuits are in the ratio

Short bread    :      Choc - chip

          2          :        3

Altogether there are 5 shares ( 2 + 3)

And now you can divide 80 by 5 to find how many biscuits contain per a share.

80 ÷ 5 = 16

In the question, they asked us to to find the number of short biscuits inside the tin.

Let us find that.

There are 2 shares for short bread biscuits.

So,

Number of short biscuits ⇒ 2 × 16

                                         ⇒ 32

Hope this helps you :-)

5 0
2 years ago
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