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3 years ago

6

a restaurant wants to study how well its salads sell the circle graph shows the sales over the past few days. if 35 of the salad

s sold were ceasar salads how many salads total did the restaurant sell

1 answer:

Schach [20]3 years ago

3 0

It’s blurry can u post it again

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In the general population, the mean IQ is 100 with a standard deviation of 15. Dr. Robbins wants to see if a class of statistics

Shalnov [3]

From the information given, a right-tailed test using the z-distribution is appropriate.

At the null hypothesis, it is <u>tested if the class of statistics students is not significantly smarter than the general population</u>, that is, their mean IQ is of at most 100, hence:

$H_0: \mu \leq 100$

At the alternative hypothesis, it is <u>tested if they are significantly smarter</u>, that is, their mean IQ is greater than 100, hence:

$H_1: \mu > 100$

We are testing if the mean is <u>greater than a value</u>, hence it is a right-tailed test.

We have the <u>standard deviation for the population</u>, hence, the z-distribution is used.

A similar problem is given at brainly.com/question/23413408

7 0

2 years ago

Find a polynomial with integer coefficients that satisfies the given conditions.

stepladder [879]

$\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)
\\\\\\
\textit{also recall that }~~~i^2=-1\\\\
-------------------------------\\\\
\begin{cases}
x=5+i\implies &x-5-i=0\\
x=5-i\implies &x-5+i=0
\end{cases}
\\\\\\
(x-5-i)(x-5+i)=\stackrel{\textit{original polynomial}}{0}
\\\\\\\
[(x-5)~~-~~i]~[(x-5)~~+~~i]=0\implies [(x-5)^2~~-~~i^2]=0
\\\\\\
(x^2-10x+25)~~-~~(-1)=0\implies x^2-10x+25+1=y
\\\\\\
x^2-10x+26=y$

8 0

2 years ago

A) Na osi liczbowej odległość między liczbami -3,2 i 2,6 jest równa ...

photoshop1234 [79]

Can you put the question in English?

8 0

3 years ago

According to one cosmological theory, there were equal amounts of the two uranium isotopes 235U and 238U at the creation of the

FromTheMoon [43]

Answer:

6 billion years.

Step-by-step explanation:

According to the decay law, the amount of the radioactive substance that decays is proportional to each instant to the amount of substance present. Let $P(t)$ be the amount of $^{235}U$ and $Q(t)$ be the amount of $^{238}U$ after $t$ years.

Then, we obtain two differential equations

$\frac{dP}{dt} = -k_1P \quad \frac{dQ}{dt} = -k_2Q$

where $k_1$ and $k_2$ are proportionality constants and the minus signs denotes decay.

Rearranging terms in the equations gives

$\frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt$

Now, the variables are separated, $P$ and $Q$ appear only on the left, and $t$ appears only on the right, so that we can integrate both sides.

$\int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt$

which yields

$\ln |P| = -k_1t + c_1 \quad \ln |Q| = -k_2t + c_2$ ,

where $c_1$ and $c_2$ are constants of integration.

By taking exponents, we obtain

$e^{\ln |P|} = e^{-k_1t + c_1} \quad e^{\ln |Q|} = e^{-k_12t + c_2}$

Hence,

$P = C_1e^{-k_1t} \quad Q = C_2e^{-k_2t}$ ,

where $C_1 := \pm e^{c_1}$ and $C_2 := \pm e^{c_2}$ .

Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition

$P(0) = Q(0) = C$

Substituting 0 for $P$ in the general solution gives

$C = P(0) = C_1 e^0 \implies C= C_1$

Similarly, we obtain $C = C_2$ and

$P = Ce^{-k_1t} \quad Q = Ce^{-k_2t}$

The relation between the decay constant $k$ and the half-life is given by

$\tau = \frac{\ln 2}{k}$

We can use this fact to determine the numeric values of the decay constants $k_1$ and $k_2$ . Thus,

$4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}$

and

$7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}$

Therefore,

$P = Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} \quad Q = Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}$

We have that

$\frac{P(t)}{Q(t)} = 137.7$

Hence,

$\frac{Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} }{Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}} = 137.7$

Solving for $t$ yields $t \approx 6 \times 10^9$ , which means that the age of the universe is about 6 billion years.

5 0

3 years ago

What is 20% of 100 in whole money

Kryger [21]

%=percent
percent means parts out of 100
20%=20/100
'of' means multiply

20% of 100 units means
20/100 times 100 units=2000/100 units=20 units

3 0

2 years ago

Read 2 more answers