Ask question

Ask question

All categories

3 years ago

12

WILL MARK BRAINLIEST A stamp collector bought 7 rare stamps for her collection. Each stamp cost $1,953. How much did the collect

or spend on the rare stamps?*

1 answer:

IceJOKER [234]3 years ago

8 0

Answer:

$13,671

Step-by-step explanation:

Each stamp cost $1,953, and the collector bought 7 of them.

1953 * 7 = 13671

The collector spent $13,671 on the stamps.

You might be interested in

Archy [21]

Answer:

Option 2 50 ≤ s ≤ 100

Option 5 She could deposit $50

Option 6 She could deposit $75

Step-by-step explanation:

Let

s -----> amount of money Layla deposit into a saving account

we know that

25%=25/100=0.25

50%=50/100=0.50

so

$s\geq 0.25*200$ -----> $s\geq \$50$

$s\leq 0.50*200$ -----> $s\leq \$100$

The compound inequality is

$\$50 \leq s\leq \$100$

<em>Verify each case</em>

case 1) 25 ≤ s ≤ 50

The statement is false

see the procedure

case 2) 50 ≤ s ≤ 100

<u>The statement is True</u>

see the procedure

case 3) s ≤ 25 or s ≥ 50

The statement is false

Because is s ≤ 100 and s ≥ 50

case 4) s ≤ 50 or s ≥ 100

The statement is false

Because is s ≤ 100 and s ≥ 50

case 5) She could deposit $50

<u>The statement is true</u>

Because the value of s satisfy the compound inequality $\$50 \leq s\leq \$100$

case 6) She could deposit $75

<u>The statement is true</u>

Because the value of s satisfy the compound inequality $\$50 \leq s\leq \$100$

3 0

3 years ago

Read 2 more answers

In one day's work at the nursery, sale and mary together potted 74 plants. If dale planted 48 of those plants how many did mary

xenn [34]

Answer:

Mary potted 26 of the plants.

Step-by-step explanation:

74 - 48 = 26

7 0

3 years ago

Read 2 more answers

Find the value of x. x√5=x^4

Kaylis [27]

Answer:

x=0, $\sqrt[6]{5}$

Decimal form,

x=0,1.30766048......

Step-by-step explanation:

x $\sqrt{5}$ = $x^{4}$

x $\sqrt{5}$ - $x^{4}$ =0

(take x as a common factor)

x( $\sqrt{5}$ - $x^{3}$ )=0

So,

x=0 and $\sqrt{5}$ - $x^{3}$ =0

for $\sqrt{5$ - $x^{3}$ =0

(add $x^{3}$ for both sides)

$\sqrt{5}$ = $x^{3}$

x= $\sqrt[3]{\sqrt[2]{5} }$ (multiply 3 by 2)

x= $\sqrt[6]{5}$

So,

x=0, $\sqrt[6]{5}$

5 0

3 years ago

P(t) = 2ť + 10<br> n(t) = 4<br> What’s the function (p+n) (t)?

Evgen [1.6K]

Answer:

$\boxed{\sf (p + n)(t) = 2t + 14}$

Given:

$\sf p(t) = 2t + 10 \\ \sf n(t) = 4$

To find:

$\sf (f + g)(x) = f(x) + g(x)$

Step-by-step explanation:

$\sf \implies(f + g)x = f(x) + g(x) \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = (2t + 10) + ( 4) \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2t + 10 + 4 \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2t + 14$

7 0

3 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

EleoNora [17]

Answer:

a) $P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

b) $P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

c) $P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

d) $E(X) = 20*0.2= 4$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=20, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We want this probability:

$P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

Part b

We want this probability:

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

Part c

We want this probability:

$P(X>3)$

We can use the complement rule and we got:

$P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

Part d

The expected value is given by:

$E(X) = np$

And replacing we got:

$E(X) = 20*0.2= 4$

3 0

3 years ago