Write 90,600,000 in scientific notation

PLEASE ANSWER FAST!! Find the 29th term of this sequence:-121,-108,-95,-82...

Julli [10]

Answer:

The 29th term is

Step-by-step explanation:

The above sequence is an arithmetic sequence

For an nth term in an arithmetic sequence

U(n) = a + ( n - 1)d

where n is the number of terms

a is the first term

d is the common difference

From the question

a = - 121

d = -108 -- 121 = - 108 + 121 = 13

Since we are finding the 29th term

n = 29

The 29th term of the sequence is

U(29) = - 121 + ( 29 - 1) 13

= -121 + 28(13)

= -121 + 364

The final answer is

Hope this helps you

4 0

3 years ago

the sweet shop has 614 pieces of candy they packed the candy into bags with 7 pieces in each bag how many did they fill how many

Nookie1986 [14]

The answer is 87.7142857

8 0

3 years ago

Read 2 more answers

If anyone knows this I will give them brainiest

statuscvo [17]

Answer:

x = 12

m(QS) = 52°

m(PD) = 152°

Step-by-step explanation:

Recall: Angle formed by two secants outside a circle = ½(the difference of the intercepted arcs)

Thus:

m<R = ½[m(PD) - m(QS)]

50° = ½[(12x + 8) - (4x + 4)] => substitution

Solve for x

Multiply both sides by 2

2*50 = (12x + 8) - (4x + 4)

100 = (12x + 8) - (4x + 4)

100 = 12x + 8 - 4x - 4 (distributive property)

Add like terms

100 = 8x + 4

100 - 4 = 8x

96 = 8x

96/8 = x

12 = x

x = 12

✔️m(QS) = 4x + 4 = 4(12) + 4 = 52°

✔️m(PD) = 12x + 8 = 12(12) + 8 = 152°

3 0

2 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0

3 years ago

What is the ratio of 65*3^17 to 65*3^14

solniwko [45]

<span>The ratio of 65*3^17 to 65*3^14 can be found as follows:

(65) (3^17) 65
--------------- = ----- * 3^(17-14) = 1 * 3^3 = 27 (answer)
(65) (3^14) 65</span>

3 0

3 years ago

Read 2 more answers