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3 years ago

43 people are going to a club. A taxi can carry 7 people how many takis do they need?

Mathematics

1 answer:

irina1246 [14]3 years ago

7 0

Answer:

They need 7 taxis

Step-by-step explanation:

43/7 is 6.14, and since their cant be .14 f a person you have to round up

You might be interested in

Determine if the expression 9y5 – 10y3 is a polynomial or not. If it is a polynomial,

Stolb23 [73]

I am going to assume that the equation was supposed to be $9y^{5}$ $-10y^{3}$

Answer:

It is a polynomial. It is a binomial and it has a degree of 5.

Step-by-step explanation:

A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables so $9y^{5}$ $-10y^{3}$ will be considered a polynomial.

A binomial is a polynomial with 2 terms.

The degree of a polynomial is the highest exponent in it which in this equation is 5.

8 0

2 years ago

What are the unit rates for 4/8 , 5/25, 36/6, 14/21, 18/30, 28/16, 56/35, 45/108, 12/42, 39/26

Artyom0805 [142]

How you find those is you just divide the rate and what ever answer you get you add a one on the bottom

Example: 1. 4/8
2.4÷8= 2
3. 2/1

5 0

3 years ago

he number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.02

Yakvenalex [24]

Answer:

a) 98.01%

b) 13.53\%

c) 27.06%

Step-by-step explanation:

Since a car has 10 square feet of plastic panel, the expected value (mean) for a car to have one flaw is 10*0.02 = 0.2

If we call P(k) the probability that a car has k flaws then, as P follows a Poisson distribution with mean 0.2,

$P(k)= \frac{0.2^ke^{-0.2}}{k!}$

In this case, we are looking for P(0)

$P(0)= \frac{0.2^0e^{-0.2}}{0!}=e^{-0.2}=0.9801=98.01\%$

So, the probability that a car has no flaws is 98.01%

Ten cars have 100 square feet of plastic panel, so now the mean is 100*0.02 = 2 flaws every ten cars.

Now P(k) is the probability that 10 cars have k flaws and

$P(k)= \frac{2^ke^{-2}}{k!}$

and

$P(0)= \frac{2^0e^{-2}}{0!}=0.1353=13.53\%$

And the probability that 10 cars have no flaws is 13.53%

Here, we are looking for P(1) with P defined as in b)

$P(1)= \frac{2^1e^{-2}}{1!}=2e^{-2}=0.2706=27.06\%$

Hence, the probability that at most one car has no flaws is 27.06%

6 0

3 years ago

Read 2 more answers

Determine the value of y, if x is -2.<br> y = x² +8

KiRa [710]

Answer:

y=x²+8

=(-2)²+8

=4+8

=12

6 0

2 years ago

The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra

Snezhnost [94]

Let $X$ denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by $X_1,\ldots,X_{36}$ , each independently and identically distributed with distribution $X_i\sim\mathcal N(26,7.2)$ .

You want to find

$\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)$

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

$\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)$

Recall that if $X\sim\mathcal N(\mu,\sigma)$ , then the sampling distribution $\overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right)$ with $n$ being the size of the sample.

Transforming to the standard normal distribution, you have

$Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}$

so that in this case,

$Z=6\dfrac{\overline X-26}{7.2}$

and the probability is equivalent to

$\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)$
$=\mathbb P(Z>1.481)\approx0.0693$

5 0

2 years ago